If the first term of a G.P. is 8 and its common ratio is -2. The 3rd term of this G.P. is :
-32
2
32
-2
Answer
Given,
First term of G.P. = 8
Common ratio = -2
By formula,
⇒ an = arn - 1
Substituting values we get :
⇒ a3 = 8 × (-2)3 - 1
⇒ a3 = 8 × (-2)2
⇒ a3 = 8 × 4
⇒ a3 = 32.
Hence, Option 3 is the correct option.
The 4th term of a G.P. is 16 and the 7th term is 128, then its common ratio is equal to :
2
-2
1
-1
Answer
Let first term of G.P. be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
4th term of a G.P. is 16.
⇒ a4 = 16
⇒ ar4 - 1 = 16
⇒ ar3 = 16 .........(1)
Given,
7th term of a G.P. is 128.
⇒ a7 = 128
⇒ ar7 - 1 = 128
⇒ ar6 = 128 .........(2)
Dividing equation (2) by (1), we get :
⇒ a r 6 a r 3 = 128 16 ⇒ r 3 = 8 ⇒ r 3 = 2 3 ⇒ r = 2. \Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{128}{16} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2. ⇒ a r 3 a r 6 = 16 128 ⇒ r 3 = 8 ⇒ r 3 = 2 3 ⇒ r = 2.
Hence, Option 1 is the correct option.
(2x + 2) and (3x + 3) are two consecutive terms of a G.P. The common ratio of the G.P. is :
2
3
3 2 \dfrac{3}{2} 2 3
2 3 \dfrac{2}{3} 3 2
Answer
By formula,
Common ratio = a n + 1 a n \dfrac{a_{n + 1}}{a_n} a n a n + 1
Given,
(2x + 2) and (3x + 3) are two consecutive terms of a G.P.
∴ Common ratio = 3 x + 3 2 x + 2 = 3 ( x + 1 ) 2 ( x + 1 ) = 3 2 . \therefore \text{Common ratio } = \dfrac{3x + 3}{2x + 2} \\[1em] = \dfrac{3(x + 1)}{2(x + 1)} \\[1em] = \dfrac{3}{2}. ∴ Common ratio = 2 x + 2 3 x + 3 = 2 ( x + 1 ) 3 ( x + 1 ) = 2 3 .
Hence, Option 3 is the correct option.
The third term of a G.P. is 3. The product of its first five terms is :
15
3 × 5 2 \dfrac{3 \times 5}{2} 2 3 × 5
35
53
Answer
Let first five terms of G.P. be
a r 2 , a r , a , a r , a r 2 \dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2 r 2 a , r a , a , a r , a r 2
Given,
3rd term of G.P. is 3.
∴ a = 3
Product of first five terms are
⇒ a r 2 × a r × a × a r × a r 2 ⇒ a 5 ⇒ 3 5 . \Rightarrow \dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2 \\[1em] \Rightarrow a^5 \\[1em] \Rightarrow 3^5. ⇒ r 2 a × r a × a × a r × a r 2 ⇒ a 5 ⇒ 3 5 .
Hence, Option 3 is the correct option.
The 8th term of a G.P. is 192 and its common ratio is 2, then the first term of the G.P. is :
3
1 3 \dfrac{1}{3} 3 1
2 3 \dfrac{2}{3} 3 2
3 2 \dfrac{3}{2} 2 3
Answer
Let first term of G.P. be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
8th term of a G.P. is 192.
⇒ a8 = 192
⇒ ar8 - 1 = 192
⇒ ar7 = 192
Substituting value of common ratio (r) = 2 in above equation, we get :
⇒ a(2)7 = 192
⇒ 128a = 192
⇒ a = 192 128 = 3 2 \dfrac{192}{128} = \dfrac{3}{2} 128 192 = 2 3
Hence, Option 4 is the correct option.
Find the 9th term of the series :
1, 4, 16, 64, ........
Answer
Since,
4 1 = 16 4 = 4 \dfrac{4}{1} = \dfrac{16}{4} = 4 1 4 = 4 16 = 4
Hence, the above sequence is a G.P. with r = 4 and a = 1.
We know that nth term of G.P.,
an = arn - 1
a9 = 1.(4)9 - 1
= (4)8
= 65536.
Hence, 9th term of the series = 65536.
Find the seventh term of the G.P. :
1, 3 , 3 , 3 3 , . . . . . . . . \sqrt{3}, 3, 3\sqrt{3}, ........ 3 , 3 , 3 3 , ........
Answer
Since,
3 1 = 3 3 3 = 3 \dfrac{\sqrt{3}}{1} = \dfrac{3\sqrt{3}}{3} = \sqrt{3} 1 3 = 3 3 3 = 3
Hence, the above sequence is a G.P. with r = 3 \sqrt{3} 3
We know that nth term of G.P.,
an = arn - 1
a7 = 1.( 3 ) 7 − 1 (\sqrt{3})^{7 - 1} ( 3 ) 7 − 1
= ( 3 ) 6 (\sqrt{3})^6 ( 3 ) 6
= 27.
Hence, 7th term of the G.P. = 27.
Find the 8th term of the sequence :
3 4 , 1 1 2 , 3 , . . . . . . . . . \dfrac{3}{4}, 1\dfrac{1}{2}, 3, ......... 4 3 , 1 2 1 , 3 , .........
Answer
Sequence = 3 4 , 3 2 , 3 , . . . . . . . . \dfrac{3}{4}, \dfrac{3}{2}, 3, ........ 4 3 , 2 3 , 3 , ........
Calculating ratio between terms,
3 2 3 4 = 3 × 4 2 × 3 = 2 , 3 3 2 = 2 \dfrac{\dfrac{3}{2}}{\dfrac{3}{4}} = \dfrac{3 \times 4}{2 \times 3} = 2, \dfrac{3}{\dfrac{3}{2}} = 2 4 3 2 3 = 2 × 3 3 × 4 = 2 , 2 3 3 = 2
Since,
3 2 3 4 = 3 3 2 = 2 \dfrac{\dfrac{3}{2}}{\dfrac{3}{4}} = \dfrac{3}{\dfrac{3}{2}} = 2 4 3 2 3 = 2 3 3 = 2
Hence, the sequence is a G.P. with r = 2 and a = 3 4 \dfrac{3}{4} 4 3
We know that nth term of G.P.,
an = arn - 1
a8 = 3 4 × ( 2 ) 8 − 1 \dfrac{3}{4} \times (2)^{8 - 1} 4 3 × ( 2 ) 8 − 1
= 3 4 × 2 7 \dfrac{3}{4} \times 2^7 4 3 × 2 7
= 3 4 × 128 \dfrac{3}{4} \times 128 4 3 × 128
= 3 × 32
= 96.
Hence, a8 = 96.
Find the next three terms of the sequence :
5 , 5 , 5 5 , . . . . . . . \sqrt{5}, 5, 5\sqrt{5}, ....... 5 , 5 , 5 5 , .......
Answer
Since,
5 5 = 5 5 5 = 5 \dfrac{5}{\sqrt{5}} = \dfrac{5\sqrt{5}}{5} = \sqrt{5} 5 5 = 5 5 5 = 5
Hence, the sequence 5 , 5 , 5 5 , . . . . . . . \sqrt{5}, 5, 5\sqrt{5}, ....... 5 , 5 , 5 5 , ....... 5 and a = 5 . \sqrt{5} \text{ and } a = \sqrt{5}. 5 and a = 5 .
Next three terms are = 4th , 5th and 6th .
We know that nth term of G.P.,
an = arn - 1
⇒ a4 = ar(4 - 1)
= ar3
= 5 ( 5 ) 3 = 5 ( 5 5 ) \sqrt{5}(\sqrt{5})^3 = \sqrt{5}(5\sqrt{5}) 5 ( 5 ) 3 = 5 ( 5 5 )
= 25.
⇒ a5 = ar(5 - 1)
= ar4
= 5 ( 5 ) 4 = 5 ( 25 ) \sqrt{5}(\sqrt{5})^4 = \sqrt{5}(25) 5 ( 5 ) 4 = 5 ( 25 )
= 25 5 25\sqrt{5} 25 5
⇒ a6 = ar(6 - 1)
= ar5
= 5 ( 5 ) 5 = 5 ( 25 5 ) \sqrt{5}(\sqrt{5})^5 = \sqrt{5}(25\sqrt{5}) 5 ( 5 ) 5 = 5 ( 25 5 )
= 125.
Hence, next three terms of the G.P. are = 25, 25√5 and 125.
Find the seventh term of the G.P. :
3 + 1 , 1 , 3 − 1 2 , \sqrt{3} + 1, 1, \dfrac{\sqrt{3} - 1}{2}, 3 + 1 , 1 , 2 3 − 1 ,
Answer
Rationalising the term, 3 − 1 2 \dfrac{\sqrt{3} - 1}{2} 2 3 − 1
⇒ 3 − 1 2 × 3 + 1 3 + 1 = 3 2 − 1 2 2 ( 3 + 1 ) = 3 − 1 2 ( 3 + 1 ) = 2 2 ( 3 + 1 ) = 1 3 + 1 . \Rightarrow \dfrac{\sqrt{3} - 1}{2} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] = \dfrac{\sqrt{3}^2 - 1^2}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{3 - 1}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{2}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{1}{\sqrt{3} + 1}. ⇒ 2 3 − 1 × 3 + 1 3 + 1 = 2 ( 3 + 1 ) 3 2 − 1 2 = 2 ( 3 + 1 ) 3 − 1 = 2 ( 3 + 1 ) 2 = 3 + 1 1 .
So, Sequence = 3 + 1 , 1 , 1 3 + 1 , \sqrt{3} + 1, 1, \dfrac{1}{\sqrt{3} + 1}, 3 + 1 , 1 , 3 + 1 1 ,
Common ratio(r) = 1 3 + 1 \dfrac{1}{\sqrt{3} + 1} 3 + 1 1
We know that nth term of G.P.,
an = arn - 1
⇒ a 7 = ( 3 + 1 ) ( 1 3 + 1 ) 7 − 1 = ( 3 + 1 ) ( 1 3 + 1 ) 6 = ( 1 3 + 1 ) 5 = ( 1 3 + 1 × 3 − 1 3 − 1 ) 5 = ( 3 − 1 3 2 − ( 1 ) 2 ) 5 = ( 3 − 1 3 − 1 ) 5 = ( 3 − 1 2 ) 5 = 1 32 ( 3 − 1 ) 5 . \Rightarrow a_7 = (\sqrt{3} + 1)\Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{7 - 1} \\[1em] = (\sqrt{3} + 1)\Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{6} \\[1em] = \Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{5} \\[1em] = \Big(\dfrac{1}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\Big)^{5} \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{\sqrt{3}^2 - (1)^2}\Big)^5 \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{3 - 1}\Big)^5 \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{2}\Big)^5 \\[1em] = \dfrac{1}{32}(\sqrt{3} - 1)^5. ⇒ a 7 = ( 3 + 1 ) ( 3 + 1 1 ) 7 − 1 = ( 3 + 1 ) ( 3 + 1 1 ) 6 = ( 3 + 1 1 ) 5 = ( 3 + 1 1 × 3 − 1 3 − 1 ) 5 = ( 3 2 − ( 1 ) 2 3 − 1 ) 5 = ( 3 − 1 3 − 1 ) 5 = ( 2 3 − 1 ) 5 = 32 1 ( 3 − 1 ) 5 .
Hence, seventh term of the G.P. = 1 32 ( 3 − 1 ) 5 . \dfrac{1}{32}(\sqrt{3} - 1)^5. 32 1 ( 3 − 1 ) 5 .
Find the next two terms of the series :
2 - 6 + 18 - 54 .............
Answer
Since,
− 6 2 = 18 − 6 \dfrac{-6}{2} = \dfrac{18}{-6} 2 − 6 = − 6 18
Hence, the above series is a G.P. with r = -3 and a = 2.
We know that nth term of G.P.,
an = arn - 1
Next two terms of the series are 5th and 6th .
⇒ a5 = ar4
= 2.(-3)4
= 2 × 81
= 162.
⇒ a6 = ar5
= 2.(-3)5
= 2 × -243
= -486.
Hence, the next two terms are 162 and -486.
Which term of the G.P. :
− 10 , 5 3 , − 5 6 , . . . . . . . . is − 5 72 ? -10, \dfrac{5}{\sqrt{3}}, -\dfrac{5}{6}, ........ \text{ is } -\dfrac{5}{72}? − 10 , 3 5 , − 6 5 , ........ is − 72 5 ?
Answer
Common ratio (r) = 5 3 − 10 = − 5 10 3 = − 1 2 3 \dfrac{\dfrac{5}{\sqrt{3}}}{-10} = -\dfrac{5}{10\sqrt{3}} = -\dfrac{1}{2\sqrt{3}} − 10 3 5 = − 10 3 5 = − 2 3 1
Let nth term of G.P. be − 5 72 -\dfrac{5}{72} − 72 5
∴ a r n − 1 = − 5 72 ⇒ − 10 × ( − 1 2 3 ) n − 1 = − 5 72 ⇒ ( − 1 2 3 ) n − 1 = − 5 72 × − 1 10 ⇒ ( − 1 2 3 ) n − 1 = 1 144 ⇒ ( − 1 2 3 ) n − 1 = ( − 1 2 3 ) 4 ⇒ n − 1 = 4 ⇒ n = 5. \therefore ar^{n - 1} = -\dfrac{5}{72} \\[1em] \Rightarrow -10 \times \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = -\dfrac{5}{72} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = -\dfrac{5}{72} \times -\dfrac{1}{10} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = \dfrac{1}{144} \\[1em] \Rightarrow \Big(-\dfrac{1}{2\sqrt{3}}\Big)^{n - 1} = \Big(-\dfrac{1}{2\sqrt{3}}\Big)^4 \\[1em] \Rightarrow n - 1 = 4 \\[1em] \Rightarrow n = 5. ∴ a r n − 1 = − 72 5 ⇒ − 10 × ( − 2 3 1 ) n − 1 = − 72 5 ⇒ ( − 2 3 1 ) n − 1 = − 72 5 × − 10 1 ⇒ ( − 2 3 1 ) n − 1 = 144 1 ⇒ ( − 2 3 1 ) n − 1 = ( − 2 3 1 ) 4 ⇒ n − 1 = 4 ⇒ n = 5.
Hence, 5th term of the G.P. is − 5 72 . -\dfrac{5}{72}. − 72 5 .
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a5 = 81
⇒ ar4 = 81 ........(i)
Also,
⇒ a2 = 24
⇒ ar = 24 ........(ii)
Dividing (i) by (ii) we get,
⇒ a r 4 a r = 81 24 ⇒ r 3 = 27 8 ⇒ r 3 = ( 3 2 ) 3 ⇒ r = 3 2 . \Rightarrow \dfrac{ar^4}{ar} = \dfrac{81}{24} \\[1em] \Rightarrow r^3 = \dfrac{27}{8} \\[1em] \Rightarrow r^3 = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow r = \dfrac{3}{2}. ⇒ a r a r 4 = 24 81 ⇒ r 3 = 8 27 ⇒ r 3 = ( 2 3 ) 3 ⇒ r = 2 3 .
Substituting value of r in (ii) we get,
⇒ a × 3 2 = 24 ⇒ a = 2 3 × 24 ⇒ a = 16. \Rightarrow a \times \dfrac{3}{2} = 24 \\[1em] \Rightarrow a = \dfrac{2}{3} \times 24 \\[1em] \Rightarrow a = 16. ⇒ a × 2 3 = 24 ⇒ a = 3 2 × 24 ⇒ a = 16.
⇒ a3 = ar2
= 16 × ( 3 2 ) 2 16 \times \Big(\dfrac{3}{2}\Big)^2 16 × ( 2 3 ) 2
= 16 × 9 4 16 \times \dfrac{9}{4} 16 × 4 9
= 4 × 9
= 36.
⇒ a4 = ar3
= 16 × ( 3 2 ) 3 16 \times \Big(\dfrac{3}{2}\Big)^3 16 × ( 2 3 ) 3
= 16 × 27 8 16 \times \dfrac{27}{8} 16 × 8 27
= 2 × 27
= 54.
G.P. = 16, 24, 36, 54, 81, ...........
Hence, G.P. = 16, 24, 36, 54, 81, ...........
Fourth and seventh terms of a G.P. are 1 18 and − 1 486 \dfrac{1}{18} \text{ and } -\dfrac{1}{486} 18 1 and − 486 1
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a4 = 1 18 \dfrac{1}{18} 18 1
⇒ ar3 = 1 18 \dfrac{1}{18} 18 1
Also,
⇒ a7 = − 1 486 -\dfrac{1}{486} − 486 1
⇒ ar6 = − 1 486 -\dfrac{1}{486} − 486 1
Dividing (ii) by (i) we get,
⇒ a r 6 a r 3 = − 1 486 1 18 ⇒ r 3 = − 18 486 ⇒ r 3 = − 1 27 ⇒ r 3 = ( − 1 3 ) 3 ⇒ r = − 1 3 . \Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{-\dfrac{1}{486}}{\dfrac{1}{18}} \\[1em] \Rightarrow r^3 = -\dfrac{18}{486} \\[1em] \Rightarrow r^3 = -\dfrac{1}{27} \\[1em] \Rightarrow r^3 = \Big(-\dfrac{1}{3}\Big)^3 \\[1em] \Rightarrow r = -\dfrac{1}{3}. ⇒ a r 3 a r 6 = 18 1 − 486 1 ⇒ r 3 = − 486 18 ⇒ r 3 = − 27 1 ⇒ r 3 = ( − 3 1 ) 3 ⇒ r = − 3 1 .
Substituting value of r in (i) we get,
⇒ a × ( − 1 3 ) 3 = 1 18 ⇒ a × − 1 27 = 1 18 ⇒ a = − 27 18 = − 3 2 . \Rightarrow a \times \Big(-\dfrac{1}{3}\Big)^3 = \dfrac{1}{18} \\[1em] \Rightarrow a \times -\dfrac{1}{27} = \dfrac{1}{18} \\[1em] \Rightarrow a = -\dfrac{27}{18} = -\dfrac{3}{2}. ⇒ a × ( − 3 1 ) 3 = 18 1 ⇒ a × − 27 1 = 18 1 ⇒ a = − 18 27 = − 2 3 .
⇒ a2 = ar
= − 3 2 × ( − 1 3 ) -\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big) − 2 3 × ( − 3 1 )
= 1 2 \dfrac{1}{2} 2 1
⇒ a3 = ar2
= − 3 2 × ( − 1 3 ) 2 -\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big)^2 − 2 3 × ( − 3 1 ) 2
= − 3 2 × 1 9 -\dfrac{3}{2} \times \dfrac{1}{9} − 2 3 × 9 1
= − 1 6 . -\dfrac{1}{6}. − 6 1 .
G.P. = − 3 2 , 1 2 , − 1 6 , 1 18 . . . . . . . . . . . -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, \dfrac{1}{18} ........... − 2 3 , 2 1 , − 6 1 , 18 1 ...........
Hence, G.P. = − 3 2 , 1 2 , − 1 6 , 1 18 . . . . . . . . . . . -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, \dfrac{1}{18} ........... − 2 3 , 2 1 , − 6 1 , 18 1 ...........
If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a = 2
⇒ a3 = 8
⇒ ar2 = 8
⇒ 2r2 = 8
⇒ r2 = 4
⇒ r =√4 = ±2
a2 = ar
Let r = -2
⇒ a2 = 2(-2) = -4
Let r = 2
⇒ a2 = 2(2) = 4.
Hence, a2 = 4 or -4.
The product of 3rd term and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a3 .a8 = 243
⇒ ar2 .ar7 = 243
⇒ a2 r9 = 243 ..........(i)
Also,
⇒ a4 = 3
⇒ ar3 = 3
⇒ a = 3 r 3 \dfrac{3}{r^3} r 3 3
Substituting value of a from (ii) in (i) we get,
⇒ ( 3 r 3 ) 2 × r 9 = 243 ⇒ 9 r 6 × r 9 = 243 ⇒ 9 r 3 = 243 ⇒ r 3 = 27 ⇒ r = 27 3 ⇒ r = 3. \Rightarrow \Big(\dfrac{3}{r^3}\Big)^2 \times r^9 = 243 \\[1em] \Rightarrow \dfrac{9}{r^6} \times r^9 = 243 \\[1em] \Rightarrow 9r^3 = 243 \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3. ⇒ ( r 3 3 ) 2 × r 9 = 243 ⇒ r 6 9 × r 9 = 243 ⇒ 9 r 3 = 243 ⇒ r 3 = 27 ⇒ r = 3 27 ⇒ r = 3.
Substituting value of r in (ii),
⇒ a = 3 r 3 = 3 3 3 = 1 3 2 = 1 9 . \Rightarrow a = \dfrac{3}{r^3} \\[1em] = \dfrac{3}{3^3} \\[1em] = \dfrac{1}{3^2} \\[1em] = \dfrac{1}{9}. ⇒ a = r 3 3 = 3 3 3 = 3 2 1 = 9 1 .
a7 = ar6
= 1 9 × 3 6 \dfrac{1}{9} \times 3^6 9 1 × 3 6
= 1 9 × 729 \dfrac{1}{9} \times 729 9 1 × 729
= 81.
Hence, 7th term = 81.
Find the geometric progression with fourth term = 54 and seventh term = 1458.
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a4 = 54
⇒ ar3 = 54 ........(i)
Also,
⇒ a7 = 1458
⇒ ar6 = 1458 ........(ii)
Dividing (ii) by (i) we get
⇒ a r 6 a r 3 = 1458 54 ⇒ r 3 = 27 ⇒ r = 27 3 ⇒ r = 3. \Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \Rightarrow r = 3. ⇒ a r 3 a r 6 = 54 1458 ⇒ r 3 = 27 ⇒ r = 3 27 ⇒ r = 3.
Substituting value of r in (i) we get,
⇒ a(3)3 = 54
⇒ 27a = 54
⇒ a = 54 27 \dfrac{54}{27} 27 54
a2 = ar
= 2.(3) = 6.
a3 = ar2
= 2.(3)2
= 2.(9) = 18.
G.P. = 2, 6, 18, 54, ......
Hence, G.P. = 2, 6, 18, 54, ......
Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a2 = 6
⇒ ar = 6 ........(i)
Also,
⇒ a5 = 9a3
⇒ ar4 = 9ar2
⇒ ar 4 ar 2 \dfrac{\text{ar}^4}{\text{ar}^2} ar 2 ar 4
⇒ r2 = 9
⇒ r = √9
⇒ r = ±3
As all terms of G.P. are positive so, r ≠ -3
∴ r = 3
Substituting r in (i),
⇒ 3a = 6
⇒ a = 2.
G.P. = a, ar, ar2 , ar3 , ......
= 2, 6, 18, 54, .......
Hence, G.P. = 2, 6, 18, 54, .......
The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a4 = 10
⇒ ar3 = 10 ........(i)
Also,
⇒ a7 = 80
⇒ ar6 = 80 .........(ii)
Dividing (ii) by (i) we get,
⇒ a r 6 a r 3 = 80 10 ⇒ r 3 = 8 ⇒ r = 8 3 ⇒ r = 2. \Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{80}{10} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} \\[1em] \Rightarrow r = 2. ⇒ a r 3 a r 6 = 10 80 ⇒ r 3 = 8 ⇒ r = 3 8 ⇒ r = 2.
Substituting r in (i) we get,
⇒ a ( 2 ) 3 = 10 ⇒ 8 a = 10 ⇒ a = 10 8 ⇒ a = 5 4 . \Rightarrow a(2)^3 = 10 \\[1em] \Rightarrow 8a = 10 \\[1em] \Rightarrow a = \dfrac{10}{8} \\[1em] \Rightarrow a = \dfrac{5}{4}. ⇒ a ( 2 ) 3 = 10 ⇒ 8 a = 10 ⇒ a = 8 10 ⇒ a = 4 5 .
Let n be no. of terms,
arn - 1 = 2560
⇒ 5 4 × ( 2 ) n − 1 = 2560 ⇒ ( 2 ) n − 1 = 4 5 × 2560 ⇒ ( 2 ) n − 1 = 4 × 512 ⇒ ( 2 ) n − 1 = 2048 ⇒ ( 2 ) n − 1 = ( 2 ) 11 ⇒ n − 1 = 11 ⇒ n = 12. \Rightarrow \dfrac{5}{4}\times (2)^{n - 1} = 2560 \\[1em] \Rightarrow (2)^{n - 1} = \dfrac{4}{5} \times 2560 \\[1em] \Rightarrow (2)^{n - 1} = 4 \times 512 \\[1em] \Rightarrow (2)^{n - 1} = 2048 \\[1em] \Rightarrow (2)^{n - 1} = (2)^{11} \\[1em] \Rightarrow n - 1 = 11 \\[1em] \Rightarrow n = 12. ⇒ 4 5 × ( 2 ) n − 1 = 2560 ⇒ ( 2 ) n − 1 = 5 4 × 2560 ⇒ ( 2 ) n − 1 = 4 × 512 ⇒ ( 2 ) n − 1 = 2048 ⇒ ( 2 ) n − 1 = ( 2 ) 11 ⇒ n − 1 = 11 ⇒ n = 12.
Hence, first term = 5 4 \dfrac{5}{4} 4 5
If the fourth and ninth terms of a G.P. are 54 and 13122 respectively, find the G.P. Also, find its general term.
Answer
Let first term of the G.P. be a and it's common ratio be r.
Given,
⇒ a4 = 54
⇒ ar3 = 54 ........(i)
Also,
⇒ a9 = 13122
⇒ ar8 = 13122 .........(ii)
Dividing (ii) by (i) we get,
⇒ a r 8 a r 3 = 13122 54 ⇒ r 5 = 243 ⇒ r 5 = 3 5 ⇒ r = 3. \Rightarrow \dfrac{ar^8}{ar^3} = \dfrac{13122}{54} \\[1em] \Rightarrow r^5 = 243 \\[1em] \Rightarrow r^5 = 3^5 \\[1em] \Rightarrow r = 3. ⇒ a r 3 a r 8 = 54 13122 ⇒ r 5 = 243 ⇒ r 5 = 3 5 ⇒ r = 3.
Substituting r in (i) we get,
⇒ a ( 3 ) 3 = 54 ⇒ 27 a = 54 ⇒ a = 2. \Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2. ⇒ a ( 3 ) 3 = 54 ⇒ 27 a = 54 ⇒ a = 2.
nth term of a G.P. = arn - 1
= 2(3)n - 1
= 2 × 3n - 1
2nd term of a G.P. = 2 × 32 - 1
= 2 × 3
= 6.
3rd term of a G.P. = 2 × 33 - 1
= 2 × 32
= 18.
G.P. = 2, 6, 18, 54, .........
Hence, G.P. = 2, 6, 18, 54, ......... and general term = 2 × 3n - 1
The fifth, eighth and eleventh terms of a geometric progression are p, q and r respectively. Show that : q2 = pr.
Answer
Let first term of the G.P. be A and it's common ratio be R.
Given,
⇒ a5 = p
⇒ AR4 = p ........(i)
Also,
⇒ a8 = q
⇒ AR7 = q .........(ii)
⇒ a11 = r
⇒ AR10 = r .........(iii)
Multiplying (i) by (iii) we get,
AR4 x AR10 = pr
⇒ A2 R14 = pr .........(iv)
Squaring eq. (ii) we get,
(AR7 )2 = q2
A2 R14 = q2 .........(v)
L.H.S. of eq. (iv) and (v) are equal so, R.H.S. will also be equal.
∴ q2 = pr.
Hence, proved that q2 = pr.
Find the seventh term from the end of the series :
2 , 2 , 2 2 , . . . . . . . . . . , 32 \sqrt{2}, 2, 2\sqrt{2}, .........., 32 2 , 2 , 2 2 , .......... , 32
Answer
Since,
2 2 = 2 2 2 = 2 \dfrac{2}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2} 2 2 = 2 2 2 = 2
So, the above sequence is a G.P. with r = 2 \sqrt{2} 2
Let there be n terms.
⇒ a r n − 1 = 32 ⇒ 2 × ( 2 ) n − 1 = 32 ⇒ 2 n − 1 + 1 = 32 ⇒ ( 2 ) n = ( 2 ) 10 ⇒ n = 10. \Rightarrow ar^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2} \times (\sqrt{2})^{n - 1} = 32 \\[1em] \Rightarrow \sqrt{2}^{n - 1 + 1} = 32 \\[1em] \Rightarrow (\sqrt{2})^n = (\sqrt{2})^{10} \\[1em] \Rightarrow n = 10. ⇒ a r n − 1 = 32 ⇒ 2 × ( 2 ) n − 1 = 32 ⇒ 2 n − 1 + 1 = 32 ⇒ ( 2 ) n = ( 2 ) 10 ⇒ n = 10.
mth term from end is (n - m + 1)th term from starting.
So, 7th term from end is (10 - 7 + 1) = 4th term from starting.
a4 = ar3
= 2 ( 2 ) 3 \sqrt{2}(\sqrt{2})^3 2 ( 2 ) 3
= 2 × 2 2 \sqrt{2} \times 2\sqrt{2} 2 × 2 2
= 4.
Hence, 7th term from end is 4.
Find the third term from the end of the G.P.
2 27 , 2 9 , 2 3 , . . . . . . . . , 162. \dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, ........, 162. 27 2 , 9 2 , 3 2 , ........ , 162.
Answer
Common ratio,
2 9 2 27 = 2 × 27 2 × 9 = 3 \dfrac{\dfrac{2}{9}}{\dfrac{2}{27}} = \dfrac{2 \times 27}{2 \times 9} = 3 27 2 9 2 = 2 × 9 2 × 27 = 3
So, the above sequence is a G.P. with r = 3.
Let there be n terms.
⇒ a r n − 1 = 162 ⇒ 2 27 × ( 3 ) n − 1 = 162 ⇒ 2 3 3 × 3 n − 1 = 162 ⇒ 3 n − 1 − 3 = 162 2 ⇒ 3 n − 4 = 81 ⇒ 3 n − 4 = 3 4 ⇒ n − 4 = 4 ⇒ n = 8. \Rightarrow ar^{n - 1} = 162 \\[1em] \Rightarrow \dfrac{2}{27} \times (3)^{n - 1} = 162 \\[1em] \Rightarrow \dfrac{2}{3^3} \times 3^{n - 1} = 162 \\[1em] \Rightarrow 3^{n - 1 - 3} = \dfrac{162}{2} \\[1em] \Rightarrow 3^{n - 4} = 81 \\[1em] \Rightarrow 3^{n - 4} = 3^4 \\[1em] \Rightarrow n - 4 = 4 \\[1em] \Rightarrow n = 8. ⇒ a r n − 1 = 162 ⇒ 27 2 × ( 3 ) n − 1 = 162 ⇒ 3 3 2 × 3 n − 1 = 162 ⇒ 3 n − 1 − 3 = 2 162 ⇒ 3 n − 4 = 81 ⇒ 3 n − 4 = 3 4 ⇒ n − 4 = 4 ⇒ n = 8.
mth term from end is (n - m + 1)th term from starting.
So, 3rd term from end is (8 - 3 + 1) = 6th term from starting.
a6 = ar5
= 2 27 × ( 3 ) 5 \dfrac{2}{27} \times (3)^5 27 2 × ( 3 ) 5
= 2 27 × 243 \dfrac{2}{27} \times 243 27 2 × 243
= 2 × 9
= 18.
Hence, 3rd term from end is 18.
For the G.P. 1 27 , 1 9 , 1 3 , . . . . . . . . . , 81 \dfrac{1}{27}, \dfrac{1}{9}, \dfrac{1}{3}, ........., 81 27 1 , 9 1 , 3 1 , ......... , 81
find the product of fourth term from the beginning and the fourth term from the end.
Answer
Common ratio,
1 9 1 27 = 27 9 = 3 \dfrac{\dfrac{1}{9}}{\dfrac{1}{27}} = \dfrac{27}{9} = 3 27 1 9 1 = 9 27 = 3
So, the above sequence is a G.P. with r = 3.
Let there be n terms.
⇒ a r n − 1 = 81 ⇒ 1 27 × ( 3 ) n − 1 = 81 ⇒ 3 n − 1 = 81 × 27 ⇒ 3 n − 1 = 3 4 × 3 3 ⇒ 3 n − 1 = 3 7 ⇒ n − 1 = 7 ⇒ n = 8. \Rightarrow ar^{n - 1} = 81 \\[1em] \Rightarrow \dfrac{1}{27} \times (3)^{n - 1} = 81 \\[1em] \Rightarrow 3^{n - 1} = 81 \times 27 \\[1em] \Rightarrow 3^{n - 1} = 3^4 \times 3^3 \\[1em] \Rightarrow 3^{n - 1} = 3^7 \\[1em] \Rightarrow n - 1 = 7 \\[1em] \Rightarrow n = 8. ⇒ a r n − 1 = 81 ⇒ 27 1 × ( 3 ) n − 1 = 81 ⇒ 3 n − 1 = 81 × 27 ⇒ 3 n − 1 = 3 4 × 3 3 ⇒ 3 n − 1 = 3 7 ⇒ n − 1 = 7 ⇒ n = 8.
a4 = ar3
= 1 27 × ( 3 ) 3 \dfrac{1}{27} \times (3)^3 27 1 × ( 3 ) 3
= 1 27 × 27 \dfrac{1}{27} \times 27 27 1 × 27
= 1.
mth term from end is (n - m + 1)th term from starting.
So, 4th term from end is (8 - 4 + 1) = 5th term from starting.
a5 = ar4
= 1 27 × ( 3 ) 4 \dfrac{1}{27} \times (3)^4 27 1 × ( 3 ) 4
= 1 27 × 81 \dfrac{1}{27} \times 81 27 1 × 81
= 3.
a4 .a5 = 1 × 3 = 3.
Hence, product of fourth term from the beginning and the fourth term from the end = 3.
If a, b and c are in G.P. and a, x, b, y, c are in A.P. prove that :
(i) 1 x + 1 y = 2 b \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b} x 1 + y 1 = b 2
(ii) a x + c y = 2. \dfrac{a}{x} + \dfrac{c}{y} = 2. x a + y c = 2.
Answer
Given,
a, b and c are in G.P.
⇒ b2 = ac .............(i)
a, x, b, y, c are in A.P.
⇒ 2x = a + b and 2y = b + c.
⇒ x = a + b 2 and y = b + c 2 x = \dfrac{a + b}{2} \text{ and } y = \dfrac{b + c}{2} x = 2 a + b and y = 2 b + c
(i) Substituting value of x and y from (ii) in L.H.S. of 1 x + 1 y = 2 b \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b} x 1 + y 1 = b 2
L.H.S . = 1 a + b 2 + 1 b + c 2 = 2 a + b + 2 b + c = 2 ( b + c ) + 2 ( a + b ) ( a + b ) ( b + c ) = 2 b + 2 c + 2 a + 2 b ( a + b ) ( b + c ) = 2 a + 2 c + 4 b a b + a c + b 2 + b c = 2 ( a + c + 2 b ) a b + b 2 + b 2 + b c from (i) = 2 ( a + c + 2 b ) b ( a + b + b + c ) = 2 ( a + c + 2 b ) b ( a + c + 2 b ) = 2 b = R.H.S. \text{L.H.S}. = \dfrac{1}{\dfrac{a + b}{2}} + \dfrac{1}{\dfrac{b + c}{2}} \\[1em] = \dfrac{2}{a + b} + \dfrac{2}{b + c} \\[1em] = \dfrac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \\[1em] = \dfrac{2b + 2c + 2a + 2b}{(a + b)(b + c)}\\[1em] = \dfrac{2a + 2c + 4b}{ab + ac + b^2 + bc} \\[1em] = \dfrac{2(a + c + 2b)}{ab + b^2 + b^2 + bc} \text{from (i)} \\[1em] = \dfrac{2(a + c + 2b)}{b(a + b + b + c)} \\[1em] = \dfrac{2(a + c + 2b)}{b(a + c + 2b)} \\[1em] = \dfrac{2}{b} = \text{R.H.S.} L.H.S . = 2 a + b 1 + 2 b + c 1 = a + b 2 + b + c 2 = ( a + b ) ( b + c ) 2 ( b + c ) + 2 ( a + b ) = ( a + b ) ( b + c ) 2 b + 2 c + 2 a + 2 b = ab + a c + b 2 + b c 2 a + 2 c + 4 b = ab + b 2 + b 2 + b c 2 ( a + c + 2 b ) from (i) = b ( a + b + b + c ) 2 ( a + c + 2 b ) = b ( a + c + 2 b ) 2 ( a + c + 2 b ) = b 2 = R.H.S.
Hence, proved that 1 x + 1 y = 2 b \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b} x 1 + y 1 = b 2
(ii) Substituting value of x and y from (ii) in L.H.S. of a x + c y = 2 \dfrac{a}{x} + \dfrac{c}{y} = 2 x a + y c = 2
L.H.S. = a a + b 2 + c b + c 2 = 2 a a + b + 2 c b + c = 2 ( a a + b + c b + c ) = 2 [ a ( b + c ) + c ( a + b ) ( a + b ) ( b + c ) ] = 2 ( a b + a c + a c + b c a b + a c + b 2 + b c ) = 2 ( a b + b 2 + b 2 + b c a b + b 2 + b 2 + b c ) . . . . . . from (i) = 2 = R.H.S. \text{L.H.S.} = \dfrac{a}{\dfrac{a + b}{2}} + \dfrac{c}{\dfrac{b + c}{2}} \\[1em] = \dfrac{2a}{a + b} + \dfrac{2c}{b + c} \\[1em] = 2\Big(\dfrac{a}{a + b} + \dfrac{c}{b + c}\Big) \\[1em] = 2\Big[\dfrac{a(b + c) + c(a + b)}{(a + b)(b + c)}\Big] \\[1em] = 2\Big(\dfrac{ab + ac + ac + bc}{ab + ac + b^2 + bc}\Big) \\[1em] = 2\Big(\dfrac{ab + b^2 + b^2 + bc}{ab + b^2 + b^2 + bc}\Big) ......\text{from (i)} \\[1em] = 2 = \text{R.H.S.} L.H.S. = 2 a + b a + 2 b + c c = a + b 2 a + b + c 2 c = 2 ( a + b a + b + c c ) = 2 [ ( a + b ) ( b + c ) a ( b + c ) + c ( a + b ) ] = 2 ( ab + a c + b 2 + b c ab + a c + a c + b c ) = 2 ( ab + b 2 + b 2 + b c ab + b 2 + b 2 + b c ) ...... from (i) = 2 = R.H.S.
Hence, proved that a x + c y = 2 \dfrac{a}{x} + \dfrac{c}{y} = 2 x a + y c = 2
If a, b and c are in A.P. and also in G.P., show that : a = b = c.
Answer
Since, a, b and c are in A.P.
⇒ 2b = a + c
⇒ b = a + c 2 \dfrac{a + c}{2} 2 a + c
Since, a, b and c are also in G.P.
⇒ b2 = ac
Substituting value of b from (i) in above equation we get,
⇒ ( a + c 2 ) 2 = a c ⇒ a 2 + c 2 + 2 a c 4 = a c ⇒ a 2 + c 2 + 2 a c = 4 a c ⇒ a 2 + c 2 − 2 a c = 0 ⇒ ( a − c ) 2 = 0 ⇒ a − c = 0 ⇒ a = c . . . . . . . ( i i ) \Rightarrow \Big(\dfrac{a + c}{2}\Big)^2 = ac \\[1em] \Rightarrow \dfrac{a^2 + c^2 + 2ac}{4} = ac \\[1em] \Rightarrow a^2 + c^2 + 2ac = 4ac \\[1em] \Rightarrow a^2 + c^2 - 2ac = 0 \\[1em] \Rightarrow (a - c)^2 = 0\\[1em] \Rightarrow a - c = 0 \\[1em] \Rightarrow a = c .......(ii) ⇒ ( 2 a + c ) 2 = a c ⇒ 4 a 2 + c 2 + 2 a c = a c ⇒ a 2 + c 2 + 2 a c = 4 a c ⇒ a 2 + c 2 − 2 a c = 0 ⇒ ( a − c ) 2 = 0 ⇒ a − c = 0 ⇒ a = c ....... ( ii )
Substituting value of a from (ii) in (i) we get,
b = c + c 2 = 2 c 2 \dfrac{c + c}{2} = \dfrac{2c}{2} 2 c + c = 2 2 c
From (ii) and (iii) we get,
a = b = c.
Hence, proved that a = b = c.
The sum of first four terms of the G.P. 2, 6, 18, ........., is :
26
80
160
52
Answer
By formula,
Common ratio = a n + 1 a n \dfrac{a_{n + 1}}{a_n} a n a n + 1
r = 6 2 \dfrac{6}{2} 2 6
By formula,
⇒ an = arn - 1
⇒ a4 = 2 × (3)4 - 1
⇒ a4 = 2 × 33
⇒ a4 = 2 × 27 = 54.
Sum of first four terms of G.P. = 2 + 6 + 18 + 54 = 80.
Hence, Option 2 is the correct option.
The 4th term of a G.P. is 54 and its 7th term is 1458, the common ratio of this G.P. is :
1 3 \dfrac{1}{3} 3 1
3
-3
− 1 3 -\dfrac{1}{3} − 3 1
Answer
Let first term of G.P. be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
4th term of a G.P. is 54.
⇒ a4 = 54
⇒ ar4 - 1 = 54
⇒ ar3 = 54 .........(1)
7th term of a G.P. is 1458.
⇒ a7 = 1458
⇒ ar7 - 1 = 1458
⇒ ar6 = 1458 .........(2)
Dividing equation (2) by (1), we get :
⇒ a r 6 a r 3 = 1458 54 ⇒ r 3 = 27 ⇒ r 3 = 3 3 ⇒ r = 3. \Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r^3 = 3^3 \\[1em] \Rightarrow r = 3. ⇒ a r 3 a r 6 = 54 1458 ⇒ r 3 = 27 ⇒ r 3 = 3 3 ⇒ r = 3.
Hence, Option 2 is the correct option.
8, x and 32 are in G.P., then the value of x is :
24
256
40
16
Answer
Given,
8, x and 32 are in G.P.
⇒ x 8 = 32 x ⇒ x 2 = 8 × 32 ⇒ x 2 = 256 ⇒ x = 256 = 16. \Rightarrow \dfrac{x}{8} = \dfrac{32}{x} \\[1em] \Rightarrow x^2 = 8 \times 32 \\[1em] \Rightarrow x^2 = 256 \\[1em] \Rightarrow x = \sqrt{256} = 16. ⇒ 8 x = x 32 ⇒ x 2 = 8 × 32 ⇒ x 2 = 256 ⇒ x = 256 = 16.
Hence, Option 4 is the correct option.
The sum of three terms (numbers) of a G.P. is 3 1 2 3\dfrac{1}{2} 3 2 1
1 2 , 1 \dfrac{1}{2}, 1 2 1 , 1
1 3 , 3 \dfrac{1}{3}, 3 3 1 , 3
1 , 1 2 1, \dfrac{1}{2} 1 , 2 1
2 , 1 2 2, \dfrac{1}{2} 2 , 2 1
Answer
Let three terms of G.P. be a r , a , a r \dfrac{a}{r}, a, ar r a , a , a r
Given,
Product of three terms of G.P. = 1
∴ a r × a × a r = 1 ⇒ a 3 = 1 ⇒ a 3 = 1 3 ⇒ a = 1. \therefore \dfrac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a^3 = 1^3 \\[1em] \Rightarrow a = 1. ∴ r a × a × a r = 1 ⇒ a 3 = 1 ⇒ a 3 = 1 3 ⇒ a = 1.
Given,
Sum of three terms of G.P. = 3 1 2 3\dfrac{1}{2} 3 2 1
⇒ a r + a + a r = 3 1 2 ⇒ 1 r + 1 + 1 ( r ) = 7 2 [ ∵ a = 1 ] ⇒ 1 r + 1 + r = 7 2 ⇒ 1 + r + r 2 r = 7 2 ⇒ 2 ( r 2 + r + 1 ) = 7 r ⇒ 2 r 2 + 2 r + 2 = 7 r ⇒ 2 r 2 + 2 r − 7 r + 2 = 0 ⇒ 2 r 2 − 5 r + 2 = 0 ⇒ 2 r 2 − 4 r − r + 2 = 0 ⇒ 2 r ( r − 2 ) − 1 ( r − 2 ) = 0 ⇒ ( 2 r − 1 ) ( r − 2 ) = 0 ⇒ 2 r − 1 = 0 or r − 2 = 0 ⇒ 2 r = 1 or r = 2 ⇒ r = 1 2 or r = 2. \Rightarrow \dfrac{a}{r} + a + ar = 3\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{r} + 1 + 1(r) = \dfrac{7}{2} \quad [\because a = 1] \\[1em] \Rightarrow \dfrac{1}{r} + 1 + r = \dfrac{7}{2} \\[1em] \Rightarrow \dfrac{1 + r + r^2}{r} = \dfrac{7}{2} \\[1em] \Rightarrow 2(r^2 + r + 1) = 7r \\[1em] \Rightarrow 2r^2 + 2r + 2 = 7r \\[1em] \Rightarrow 2r^2 + 2r - 7r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 5r + 2 = 0 \\[1em] \Rightarrow 2r^2 - 4r - r + 2 = 0 \\[1em] \Rightarrow 2r(r - 2) - 1(r - 2) = 0 \\[1em] \Rightarrow (2r - 1)(r - 2) = 0 \\[1em] \Rightarrow 2r - 1 = 0 \text{ or } r - 2 = 0 \\[1em] \Rightarrow 2r = 1 \text{ or } r = 2 \\[1em] \Rightarrow r = \dfrac{1}{2} \text{ or } r = 2. ⇒ r a + a + a r = 3 2 1 ⇒ r 1 + 1 + 1 ( r ) = 2 7 [ ∵ a = 1 ] ⇒ r 1 + 1 + r = 2 7 ⇒ r 1 + r + r 2 = 2 7 ⇒ 2 ( r 2 + r + 1 ) = 7 r ⇒ 2 r 2 + 2 r + 2 = 7 r ⇒ 2 r 2 + 2 r − 7 r + 2 = 0 ⇒ 2 r 2 − 5 r + 2 = 0 ⇒ 2 r 2 − 4 r − r + 2 = 0 ⇒ 2 r ( r − 2 ) − 1 ( r − 2 ) = 0 ⇒ ( 2 r − 1 ) ( r − 2 ) = 0 ⇒ 2 r − 1 = 0 or r − 2 = 0 ⇒ 2 r = 1 or r = 2 ⇒ r = 2 1 or r = 2.
Let r = 1 2 \dfrac{1}{2} 2 1
Terms :
⇒ a r \dfrac{a}{r} r a
⇒ 1 1 2 , 1 , 1 × 1 2 \dfrac{1}{\dfrac{1}{2}}, 1, 1 \times \dfrac{1}{2} 2 1 1 , 1 , 1 × 2 1
⇒ 2, 1, 1 2 \dfrac{1}{2} 2 1
Let r = 2
Terms :
⇒ a r \dfrac{a}{r} r a
⇒ 1 2 , 1 , 1 × 2 \dfrac{1}{2}, 1, 1 \times 2 2 1 , 1 , 1 × 2
⇒ 1 2 \dfrac{1}{2} 2 1
Hence, Option 1 is the correct option.
The sum of 20 terms of the G.P. 10, 20, 40, ...... is :
10(219 - 1)
10(221 - 1)
10(220 - 1)
none of these
Answer
By formula,
Common ratio = a n + 1 a n \dfrac{a_{n + 1}}{a_n} a n a n + 1
Given,
G.P. = 10, 20, 40, ......
a = 10 and r = 20 10 \dfrac{20}{10} 10 20
We know that,
If | r | > 1
Sum of n terms of G.P. (Sn ) = a ( r n − 1 ) ( r − 1 ) \dfrac{a(r^n - 1)}{(r - 1)} ( r − 1 ) a ( r n − 1 )
Substituting values we get :
⇒ S 20 = 10 ( 2 20 − 1 ) 2 − 1 = 10 ( 2 20 − 1 ) 1 = 10 ( 2 20 − 1 ) . \Rightarrow S_{20} = \dfrac{10(2^{20} - 1)}{2 - 1} \\[1em] = \dfrac{10(2^{20} - 1)}{1} \\[1em] = 10(2^{20} - 1). ⇒ S 20 = 2 − 1 10 ( 2 20 − 1 ) = 1 10 ( 2 20 − 1 ) = 10 ( 2 20 − 1 ) .
Hence, Option 3 is the correct option.
Find the sum of G.P. :
1 + 3 + 9 + 27 + ........ to 12 terms
Answer
Common ratio (r) = 3 1 \dfrac{3}{1} 1 3
S = a ( r n − 1 ) ( r − 1 ) . . . . . . . . . . ( A s ∣ r ∣ > 1 ) = 1 ( 3 12 − 1 ) 3 − 1 = 531441 − 1 2 = 531440 2 = 265720. S = \dfrac{a(r^n - 1)}{(r - 1)} ..........(As |r| \gt 1)\\[1em] = \dfrac{1(3^{12} - 1)}{3 - 1} \\[1em] = \dfrac{531441 - 1}{2} \\[1em] = \dfrac{531440}{2} \\[1em] = 265720. S = ( r − 1 ) a ( r n − 1 ) .......... ( A s ∣ r ∣ > 1 ) = 3 − 1 1 ( 3 12 − 1 ) = 2 531441 − 1 = 2 531440 = 265720.
Hence, sum = 265720.
Find the sum of G.P. :
0.3 + 0.03 + 0.003 + 0.0003 + ...... to 8 terms.
Answer
Common ratio (r) = 0.03 0.3 \dfrac{0.03}{0.3} 0.3 0.03
S = a ( 1 − r n ) ( 1 − r ) . . . . . . . . . . ( A s ∣ r ∣ < 1 ) = 0.3 [ 1 − ( 0.1 ) 8 ] 1 − 0.1 = 0.3 [ 1 − ( 1 10 ) 8 ] 0.9 = 1 3 ( 1 − 1 10 8 ) . S = \dfrac{a(1 - r^n)}{(1 - r)} ..........(As |r| \lt 1)\\[1em] = \dfrac{0.3\Big[1 - (0.1)^8\Big]}{1 - 0.1} \\[1em] = \dfrac{0.3\Big[1 - \Big(\dfrac{1}{10}\Big)^8\Big]}{0.9} \\[1em] = \dfrac{1}{3}\Big(1 - \dfrac{1}{10^8}\Big). S = ( 1 − r ) a ( 1 − r n ) .......... ( A s ∣ r ∣ < 1 ) = 1 − 0.1 0.3 [ 1 − ( 0.1 ) 8 ] = 0.9 0.3 [ 1 − ( 10 1 ) 8 ] = 3 1 ( 1 − 1 0 8 1 ) .
Hence, sum = 1 3 ( 1 − 1 10 8 ) \dfrac{1}{3}\Big(1 - \dfrac{1}{10^8}\Big) 3 1 ( 1 − 1 0 8 1 )
Find the sum of G.P. :
1 − 1 2 + 1 4 − 1 8 + . . . . . . . 1 - \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} + ....... 1 − 2 1 + 4 1 − 8 1 + .......
Answer
Common ratio (r) = − 1 2 1 = − 1 2 \dfrac{-\dfrac{1}{2}}{1} = -\dfrac{1}{2} 1 − 2 1 = − 2 1
S = a ( 1 − r n ) ( 1 − r ) . . . . . . . . . . ( A s ∣ r ∣ < 1 ) = 1 [ 1 − ( − 1 2 ) 9 ] 1 − ( − 1 2 ) = [ 1 + ( 1 2 9 ) ] 1 + 1 2 = ( 1 + 1 2 9 ) 3 2 = 2 3 ( 1 + 1 2 9 ) S = \dfrac{a(1 - r^n)}{(1 - r)} ..........(As |r| \lt 1) \\[1em] = \dfrac{1\Big[1 - \Big(-\dfrac{1}{2}\Big)^9\Big]}{1 - \Big(-\dfrac{1}{2}\Big)} \\[1em] = \dfrac{\Big[1 +\Big(\dfrac{1}{2^9}\Big)\Big]}{1 + \dfrac{1}{2}} \\[1em] = \dfrac{\Big(1 + \dfrac{1}{2^9}\Big)}{\dfrac{3}{2}} \\[1em] = \dfrac{2}{3}\Big(1 + \dfrac{1}{2^9}\Big) S = ( 1 − r ) a ( 1 − r n ) .......... ( A s ∣ r ∣ < 1 ) = 1 − ( − 2 1 ) 1 [ 1 − ( − 2 1 ) 9 ] = 1 + 2 1 [ 1 + ( 2 9 1 ) ] = 2 3 ( 1 + 2 9 1 ) = 3 2 ( 1 + 2 9 1 )
Hence, sum = 2 3 ( 1 + 1 2 9 ) \dfrac{2}{3}\Big(1 + \dfrac{1}{2^9}\Big) 3 2 ( 1 + 2 9 1 )
How many terms of the geometric progression 1 + 4 + 16 + 64 + ........ must be added to get sum equal to 5461 ?
Answer
Let n terms be added.
Common ratio = 4 1 \dfrac{4}{1} 1 4
⇒ S = a ( r n − 1 ) ( r − 1 ) . . . . . . . . . . ( A s ∣ r ∣ > 1 ) ⇒ 5461 = 1 [ ( 4 ) n − 1 ] 4 − 1 ⇒ 4 n − 1 3 = 5461 ⇒ 4 n − 1 = 16383 ⇒ 4 n = 16384 ⇒ 4 n = 4 7 ⇒ n = 7. \Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} ..........(As |r| \gt 1)\\[1em] \Rightarrow 5461 = \dfrac{1[(4)^n - 1]}{4 - 1} \\[1em] \Rightarrow \dfrac{4^n - 1}{3} = 5461 \\[1em] \Rightarrow 4^n - 1 = 16383 \\[1em] \Rightarrow 4^n = 16384 \\[1em] \Rightarrow 4^n = 4^7 \\[1em] \Rightarrow n = 7. ⇒ S = ( r − 1 ) a ( r n − 1 ) .......... ( A s ∣ r ∣ > 1 ) ⇒ 5461 = 4 − 1 1 [( 4 ) n − 1 ] ⇒ 3 4 n − 1 = 5461 ⇒ 4 n − 1 = 16383 ⇒ 4 n = 16384 ⇒ 4 n = 4 7 ⇒ n = 7.
Hence, 7 terms must be added to get a sum of 5461.
The first term of a G.P. is 27 and its 8th term is 1 81 \dfrac{1}{81} 81 1
Answer
Given , a = 27 and a8 = 1 81 \dfrac{1}{81} 81 1
∴ a r 7 = 1 81 ⇒ 27 r 7 = 1 81 ⇒ r 7 = 1 81 × 27 ⇒ r 7 = 1 3 4 × 3 3 ⇒ r 7 = ( 1 3 ) 7 ⇒ r = 1 3 . \therefore ar^7 = \dfrac{1}{81} \\[1em] \Rightarrow 27r^7 = \dfrac{1}{81} \\[1em] \Rightarrow r^7 = \dfrac{1}{81 \times 27} \\[1em] \Rightarrow r^7 = \dfrac{1}{3^4 \times 3^3} \\[1em] \Rightarrow r^7 = \Big(\dfrac{1}{3}\Big)^7 \\[1em] \Rightarrow r = \dfrac{1}{3}. ∴ a r 7 = 81 1 ⇒ 27 r 7 = 81 1 ⇒ r 7 = 81 × 27 1 ⇒ r 7 = 3 4 × 3 3 1 ⇒ r 7 = ( 3 1 ) 7 ⇒ r = 3 1 .
Since, r < 1
S = a ( 1 − r n ) ( 1 − r ) = 27 [ 1 − ( 1 3 ) 10 ] 1 − 1 3 = 27 ( 1 − 1 3 10 ) 2 3 = 81 2 ( 1 − 1 3 10 ) . S = \dfrac{a(1 - r^n)}{(1 - r)} \\[1em] = \dfrac{27\Big[1 - \Big(\dfrac{1}{3}\Big)^{10}\Big]}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{27\Big(1 - \dfrac{1}{3^{10}}\Big)}{\dfrac{2}{3}} \\[1em] = \dfrac{81}{2}\Big(1 - \dfrac{1}{3^{10}}\Big). S = ( 1 − r ) a ( 1 − r n ) = 1 − 3 1 27 [ 1 − ( 3 1 ) 10 ] = 3 2 27 ( 1 − 3 10 1 ) = 2 81 ( 1 − 3 10 1 ) .
Hence, sum upto 10 terms = 81 2 ( 1 − 1 3 10 ) . \dfrac{81}{2}\Big(1 - \dfrac{1}{3^{10}}\Big). 2 81 ( 1 − 3 10 1 ) .
A boy spends ₹ 10 on first day, ₹ 20 on second day, ₹ 40 on third day and so on. Find how much in all, will he spend in 12 days?
Answer
G.P. formed = 10 + 20 + 40 + ........
Common ratio (r) = 20 10 \dfrac{20}{10} 10 20
⇒ S = a ( r n − 1 ) ( r − 1 ) . . . . . . . . . . ( A s ∣ r ∣ > 1 ) = 10 ( 2 12 − 1 ) 2 − 1 = 10 ( 2 12 − 1 ) . \Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} ..........(As |r| \gt 1)\\[1em] = \dfrac{10(2^{12} - 1)}{2 - 1} \\[1em] = 10(2^{12} - 1). ⇒ S = ( r − 1 ) a ( r n − 1 ) .......... ( A s ∣ r ∣ > 1 ) = 2 − 1 10 ( 2 12 − 1 ) = 10 ( 2 12 − 1 ) .
Hence, ₹10(212 - 1).
The 4th and the 7th terms of a G.P. are 1 27 and 1 729 \dfrac{1}{27} \text{ and } \dfrac{1}{729} 27 1 and 729 1
Answer
Given,
a 4 = 1 27 a_4 = \dfrac{1}{27} a 4 = 27 1
a r 3 = 1 27 ar^3 = \dfrac{1}{27} a r 3 = 27 1
a 7 = 1 729 a_7 = \dfrac{1}{729} a 7 = 729 1
a r 6 = 1 729 ar^6 = \dfrac{1}{729} a r 6 = 729 1
Dividing (ii) by (i) we get,
⇒ a r 6 a r 3 = 1 729 1 27 ⇒ r 3 = 27 729 ⇒ r 3 = 1 27 ⇒ r = 1 3 . \Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{\dfrac{1}{729}}{\dfrac{1}{27}} \\[1em] \Rightarrow r^3 = \dfrac{27}{729} \\[1em] \Rightarrow r^3 = \dfrac{1}{27} \\[1em] \Rightarrow r = \dfrac{1}{3}. ⇒ a r 3 a r 6 = 27 1 729 1 ⇒ r 3 = 729 27 ⇒ r 3 = 27 1 ⇒ r = 3 1 .
Substituting value of r in (i) we get,
⇒ a ( 1 3 ) 3 = 1 27 ⇒ a × 1 27 = 1 27 ⇒ a = 1. \Rightarrow a\Big(\dfrac{1}{3}\Big)^3 = \dfrac{1}{27} \\[1em] \Rightarrow a \times \dfrac{1}{27} = \dfrac{1}{27} \\[1em] \Rightarrow a = 1. ⇒ a ( 3 1 ) 3 = 27 1 ⇒ a × 27 1 = 27 1 ⇒ a = 1.
Since, r < 1
S = a ( 1 − r n ) ( 1 − r ) = 1 [ 1 − ( 1 3 ) n ] 1 − 1 3 = [ 1 − ( 1 3 ) n ] 2 3 = 3 2 [ 1 − ( 1 3 ) n ] = 3 2 ( 1 − 1 3 n ) S = \dfrac{a(1 - r^n)}{(1 - r)} \\[1em] = \dfrac{1\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big]}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big]}{\dfrac{2}{3}} \\[1em] = \dfrac{3}{2}\Big[1 - \Big(\dfrac{1}{3}\Big)^n \Big] \\[1em] = \dfrac{3}{2}\Big(1 - \dfrac{1}{3^n} \Big) S = ( 1 − r ) a ( 1 − r n ) = 1 − 3 1 1 [ 1 − ( 3 1 ) n ] = 3 2 [ 1 − ( 3 1 ) n ] = 2 3 [ 1 − ( 3 1 ) n ] = 2 3 ( 1 − 3 n 1 )
Hence, sum = 3 2 ( 1 − 1 3 n ) . \dfrac{3}{2}\Big(1 - \dfrac{1}{3^n} \Big). 2 3 ( 1 − 3 n 1 ) .
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.
Answer
Let nth term be the last term.
⇒ arn - 1 = 486
⇒ a(3)n - 1 = 486
⇒ a.3n .3-1 = 486
⇒ 3 n = 486 × 3 a = 1458 a 3^n = \dfrac{486 \times 3}{a} = \dfrac{1458}{a} 3 n = a 486 × 3 = a 1458
Since, r > 1
⇒ S = a ( r n − 1 ) ( r − 1 ) ⇒ 728 = a × ( 3 n − 1 ) 3 − 1 ⇒ 728 = a × ( 1458 a − 1 ) 2 ⇒ 728 = a × ( 1458 − a ) 2 a ⇒ 1458 − a 2 = 728 ⇒ 1458 − a = 1456 ⇒ a = 1458 − 1456 ⇒ a = 2. \Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow 728 = \dfrac{a \times (3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \dfrac{a \times \Big(\dfrac{1458}{a} - 1\Big)}{2} \\[1em] \Rightarrow 728 = \dfrac{a \times (1458 - a)}{2a} \\[1em] \Rightarrow \dfrac{1458 - a}{2} = 728 \\[1em] \Rightarrow 1458 - a = 1456 \\[1em] \Rightarrow a = 1458 - 1456 \\[1em] \Rightarrow a = 2. ⇒ S = ( r − 1 ) a ( r n − 1 ) ⇒ 728 = 3 − 1 a × ( 3 n − 1 ) ⇒ 728 = 2 a × ( a 1458 − 1 ) ⇒ 728 = 2 a a × ( 1458 − a ) ⇒ 2 1458 − a = 728 ⇒ 1458 − a = 1456 ⇒ a = 1458 − 1456 ⇒ a = 2.
Hence, first term = 2.
Find the sum of G.P. : 3, 6, 12, ......., 1536.
Answer
Common ratio = 6 3 \dfrac{6}{3} 3 6
Let 1536 be nth term
∴ a r n − 1 = 1536 ⇒ 3. ( 2 ) n − 1 = 1536 ⇒ ( 2 ) n − 1 = 512 ⇒ ( 2 ) n − 1 = ( 2 ) 9 ⇒ n − 1 = 9 ⇒ n = 10. \therefore ar^{n - 1} = 1536 \\[1em] \Rightarrow 3.(2)^{n - 1} = 1536 \\[1em] \Rightarrow (2)^{n - 1} = 512 \\[1em] \Rightarrow (2)^{n - 1} = (2)^{9} \\[1em] \Rightarrow n - 1 = 9 \\[1em] \Rightarrow n = 10. ∴ a r n − 1 = 1536 ⇒ 3. ( 2 ) n − 1 = 1536 ⇒ ( 2 ) n − 1 = 512 ⇒ ( 2 ) n − 1 = ( 2 ) 9 ⇒ n − 1 = 9 ⇒ n = 10.
Since, r > 1
S = a ( r n − 1 ) ( r − 1 ) = 3 × ( 2 10 − 1 ) 2 − 1 = 3 ( 2 10 − 1 ) = 3 ( 1024 − 1 ) = 3 × 1023 = 3069. S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] = \dfrac{3 \times (2^{10} - 1)}{2 - 1} \\[1em] = 3(2^{10} - 1) \\[1em] = 3(1024 - 1) \\[1em] = 3 \times 1023 \\[1em] = 3069. S = ( r − 1 ) a ( r n − 1 ) = 2 − 1 3 × ( 2 10 − 1 ) = 3 ( 2 10 − 1 ) = 3 ( 1024 − 1 ) = 3 × 1023 = 3069.
Hence, sum = 3069.
How many terms of the series 2 + 6 + 18 + ...... must be taken to make the sum equal to 728?
Answer
Common ratio = 6 2 \dfrac{6}{2} 2 6
Let n be no. of terms taken.
Since, r > 1
⇒ S = a ( r n − 1 ) ( r − 1 ) ⇒ 728 = 2 × ( 3 n − 1 ) 3 − 1 ⇒ 728 = 2 ( 3 n − 1 ) 2 ⇒ 3 n − 1 = 728 ⇒ 3 n = 729 ⇒ 3 n = 3 6 ⇒ n = 6. \Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow 728 = \dfrac{2 \times (3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \dfrac{2(3^n - 1)}{2} \\[1em] \Rightarrow 3^n - 1 = 728 \\[1em] \Rightarrow 3^n = 729 \\[1em] \Rightarrow 3^n = 3^6 \\[1em] \Rightarrow n = 6. ⇒ S = ( r − 1 ) a ( r n − 1 ) ⇒ 728 = 3 − 1 2 × ( 3 n − 1 ) ⇒ 728 = 2 2 ( 3 n − 1 ) ⇒ 3 n − 1 = 728 ⇒ 3 n = 729 ⇒ 3 n = 3 6 ⇒ n = 6.
Hence, 6 terms must be taken to make the sum equal to 728.
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152. Find its common ratio.
Answer
Given,
⇒ S 3 S 6 = 125 152 ⇒ a ( r 3 − 1 ) r − 1 a ( r 6 − 1 ) r − 1 = 125 152 ⇒ r 3 − 1 r 6 − 1 = 125 152 ⇒ r 3 − 1 ( r 3 − 1 ) ( r 3 + 1 ) = 125 152 ⇒ 1 r 3 + 1 = 125 152 ⇒ r 3 + 1 = 152 125 ⇒ r 3 = 152 125 − 1 ⇒ r 3 = 152 − 125 125 ⇒ r 3 = 27 125 ⇒ r 3 = ( 3 5 ) 3 ⇒ r = 3 5 . \Rightarrow \dfrac{S_3}{S_6} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{\dfrac{a(r^3 - 1)}{r - 1}}{\dfrac{a(r^6 - 1)}{r - 1}} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{r^3 - 1}{r^6 - 1} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \dfrac{125}{152} \\[1em] \Rightarrow \dfrac{1}{r^3 + 1} = \dfrac{125}{152} \\[1em] \Rightarrow r^3 + 1 = \dfrac{152}{125} \\[1em] \Rightarrow r^3 = \dfrac{152}{125} - 1 \\[1em] \Rightarrow r^3 = \dfrac{152 - 125}{125} \\[1em] \Rightarrow r^3 = \dfrac{27}{125} \\[1em] \Rightarrow r^3 = \Big(\dfrac{3}{5}\Big)^3 \\[1em] \Rightarrow r = \dfrac{3}{5}. ⇒ S 6 S 3 = 152 125 ⇒ r − 1 a ( r 6 − 1 ) r − 1 a ( r 3 − 1 ) = 152 125 ⇒ r 6 − 1 r 3 − 1 = 152 125 ⇒ ( r 3 − 1 ) ( r 3 + 1 ) r 3 − 1 = 152 125 ⇒ r 3 + 1 1 = 152 125 ⇒ r 3 + 1 = 125 152 ⇒ r 3 = 125 152 − 1 ⇒ r 3 = 125 152 − 125 ⇒ r 3 = 125 27 ⇒ r 3 = ( 5 3 ) 3 ⇒ r = 5 3 .
Hence, common ratio = 3 5 \dfrac{3}{5} 5 3
If the sum of 1 + 2 + 22 + ........ + 2n - 1 is 255, find the value of n.
Answer
Common ratio = 2.
Since, |r| > 1
⇒ S = a ( r n − 1 ) ( r − 1 ) ⇒ 255 = 1 ( 2 n − 1 ) ( 2 − 1 ) ⇒ 2 n − 1 = 255 ⇒ 2 n = 256 ⇒ 2 n = 2 8 ⇒ n = 8. \Rightarrow S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] \Rightarrow 255 = \dfrac{1(2^n - 1)}{(2 - 1)} \\[1em] \Rightarrow 2^n - 1 = 255 \\[1em] \Rightarrow 2^n = 256 \\[1em] \Rightarrow 2^n = 2^8 \\[1em] \Rightarrow n = 8. ⇒ S = ( r − 1 ) a ( r n − 1 ) ⇒ 255 = ( 2 − 1 ) 1 ( 2 n − 1 ) ⇒ 2 n − 1 = 255 ⇒ 2 n = 256 ⇒ 2 n = 2 8 ⇒ n = 8.
Hence, n = 8.
The sum of three numbers in G.P. is 39 10 \dfrac{39}{10} 10 39
Answer
Let the numbers be a r , a , a r \dfrac{a}{r}, a, ar r a , a , a r
Given,
Product = 1.
∴ a r × a × a r = 1 ⇒ a 3 = 1 ⇒ a = 1. \therefore \dfrac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a = 1. ∴ r a × a × a r = 1 ⇒ a 3 = 1 ⇒ a = 1.
Sum = 39 10 \dfrac{39}{10} 10 39
∴ a r + a + a r = 39 10 ⇒ 1 r + 1 + r = 39 10 ⇒ 1 + r + r 2 r = 39 10 ⇒ 10 ( 1 + r + r 2 ) = 39 r ⇒ 10 + 10 r + 10 r 2 = 39 r ⇒ 10 r 2 − 29 r + 10 = 0 ⇒ 10 r 2 − 25 r − 4 r + 10 = 0 ⇒ 5 r ( 2 r − 5 ) − 2 ( 2 r − 5 ) = 0 ⇒ ( 5 r − 2 ) ( 2 r − 5 ) = 0 ⇒ 5 r − 2 = 0 or 2 r − 5 = 0 ⇒ r = 2 5 or r = 5 2 \therefore \dfrac{a}{r} + a + ar = \dfrac{39}{10} \\[1em] \Rightarrow \dfrac{1}{r} + 1 + r = \dfrac{39}{10} \\[1em] \Rightarrow \dfrac{1 + r + r^2}{r} = \dfrac{39}{10} \\[1em] \Rightarrow 10(1 + r + r^2) = 39r \\[1em] \Rightarrow 10 + 10r + 10r^2 = 39r \\[1em] \Rightarrow 10r^2 - 29r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \\[1em] \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \\[1em] \Rightarrow (5r - 2)(2r - 5) = 0 \\[1em] \Rightarrow 5r - 2 = 0 \text{ or } 2r - 5 = 0 \\[1em] \Rightarrow r = \dfrac{2}{5} \text{ or } r = \dfrac{5}{2} ∴ r a + a + a r = 10 39 ⇒ r 1 + 1 + r = 10 39 ⇒ r 1 + r + r 2 = 10 39 ⇒ 10 ( 1 + r + r 2 ) = 39 r ⇒ 10 + 10 r + 10 r 2 = 39 r ⇒ 10 r 2 − 29 r + 10 = 0 ⇒ 10 r 2 − 25 r − 4 r + 10 = 0 ⇒ 5 r ( 2 r − 5 ) − 2 ( 2 r − 5 ) = 0 ⇒ ( 5 r − 2 ) ( 2 r − 5 ) = 0 ⇒ 5 r − 2 = 0 or 2 r − 5 = 0 ⇒ r = 5 2 or r = 2 5
Let r = 2 5 \dfrac{2}{5} 5 2
Numbers = a r = 1 2 5 = 5 2 \dfrac{a}{r} = \dfrac{1}{\dfrac{2}{5}} = \dfrac{5}{2} r a = 5 2 1 = 2 5
a = 1
ar = 1 × 2 5 = 2 5 1 \times \dfrac{2}{5} =\dfrac{2}{5} 1 × 5 2 = 5 2
Let r = 5 2 \dfrac{5}{2} 2 5
Numbers = a r = 1 5 2 = 2 5 \dfrac{a}{r} = \dfrac{1}{\dfrac{5}{2}} = \dfrac{2}{5} r a = 2 5 1 = 5 2
a = 1
ar = 1 × 5 2 = 5 2 1 \times \dfrac{5}{2} =\dfrac{5}{2} 1 × 2 5 = 2 5
Hence, numbers = 5 2 , 1 , 2 5 or 2 5 , 1 , 5 2 \dfrac{5}{2}, 1, \dfrac{2}{5} \text{ or } \dfrac{2}{5}, 1, \dfrac{5}{2} 2 5 , 1 , 5 2 or 5 2 , 1 , 2 5
The first term of a G.P. is -3 and the square of the second term is equal to its 4th term. Find its 7th term.
Answer
Given, a = -3 and,
(a2 )2 = a4
(ar)2 = ar3
(-3r)2 = -3r3
9r2 = -3r3
r = -3.
a7 = ar6 = (-3)(-3)6
= -3 × 729
= -2187.
Hence, 7th term = -2187.
Find the 5th term of the G.P. 5 2 , 1 , . . . . . . . \dfrac{5}{2}, 1, ....... 2 5 , 1 , .......
Answer
Common ratio = 1 5 2 = 2 5 \dfrac{1}{\dfrac{5}{2}} = \dfrac{2}{5} 2 5 1 = 5 2
a5 = ar4
= 5 2 × ( 2 5 ) 4 \dfrac{5}{2} \times \Big(\dfrac{2}{5}\Big)^4 2 5 × ( 5 2 ) 4
= 5 2 × 16 625 \dfrac{5}{2} \times \dfrac{16}{625} 2 5 × 625 16
= 8 125 \dfrac{8}{125} 125 8
Hence, 5th term = 8 125 \dfrac{8}{125} 125 8
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Answer
Given,
⇒ a = 125
⇒ a2 = 25
⇒ ar = 25
⇒ 125r = 25
⇒ r = 25 125 = 1 5 \dfrac{25}{125} = \dfrac{1}{5} 125 25 = 5 1
a5 = ar4
= 125 × ( 1 5 ) 4 125 \times \Big(\dfrac{1}{5}\Big)^4 125 × ( 5 1 ) 4
= 125 × 1 625 125 \times \dfrac{1}{625} 125 × 625 1
= 1 5 . \dfrac{1}{5}. 5 1 .
a6 = ar5
= 125 × ( 1 5 ) 5 125 \times \Big(\dfrac{1}{5}\Big)^5 125 × ( 5 1 ) 5
= 125 × 1 3125 125 \times \dfrac{1}{3125} 125 × 3125 1
= 1 25 . \dfrac{1}{25}. 25 1 .
Hence, a 5 = 1 5 and a 6 = 1 25 . a_5 = \dfrac{1}{5} \text{ and } a_6 = \dfrac{1}{25}. a 5 = 5 1 and a 6 = 25 1 .
Find the sum of the sequence − 1 3 , 1 , − 3 , 9 , . . . . . -\dfrac{1}{3}, 1, -3, 9, ..... − 3 1 , 1 , − 3 , 9 , .....
Answer
Common ratio = 1 − 1 3 = − 3 \dfrac{1}{-\dfrac{1}{3}} = -3 − 3 1 1 = − 3
Since, |r| > 1
S = a ( r n − 1 ) ( r − 1 ) = − 1 3 [ ( − 3 ) 8 − 1 ] − 3 − 1 = − 1 3 × [ ( − 1 ) 8 ( 3 ) 8 − 1 ] − 4 = 1 12 ( 3 8 − 1 ) . S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] = \dfrac{-\dfrac{1}{3}[(-3)^8 - 1]}{-3 - 1} \\[1em] = -\dfrac{1}{3} \times \dfrac{[(-1)^8(3)^8 - 1]}{-4} \\[1em] = \dfrac{1}{12}(3^8 - 1). S = ( r − 1 ) a ( r n − 1 ) = − 3 − 1 − 3 1 [( − 3 ) 8 − 1 ] = − 3 1 × − 4 [( − 1 ) 8 ( 3 ) 8 − 1 ] = 12 1 ( 3 8 − 1 ) .
Hence, sum = 1 12 ( 3 8 − 1 ) . \dfrac{1}{12}(3^8 - 1). 12 1 ( 3 8 − 1 ) .
The first term of a G.P. is 27. If the 8th term be 1 81 \dfrac{1}{81} 81 1
Answer
Given,
⇒ a = 27
⇒ a8 = ar7 = 1 81 \dfrac{1}{81} 81 1
⇒ 27r7 = 1 81 \dfrac{1}{81} 81 1
⇒ r7 = 1 81 × 27 \dfrac{1}{81 \times 27} 81 × 27 1
⇒ r7 = 1 3 7 \dfrac{1}{3^7} 3 7 1
⇒ r7 = ( 1 3 ) 7 \Big(\dfrac{1}{3}\Big)^7 ( 3 1 ) 7
⇒ r = 1 3 \dfrac{1}{3} 3 1
Since, r < 1
S = a ( 1 − r n ) ( 1 − r ) = 27 [ 1 − ( 1 3 ) 10 ] 1 − 1 3 = 27 [ 1 − ( 1 3 ) 10 ] 2 3 = 81 2 [ 1 − ( 1 3 ) 10 ] = 81 2 ( 1 − 3 − 10 ) . S = \dfrac{a(1 - r^n)}{(1 - r)} \\[1em] = \dfrac{27\Big[1 - \Big(\dfrac{1}{3}\Big)^{10}\Big]}{1 - \dfrac{1}{3}} \\[1em] = \dfrac{27\Big[1 - \Big(\dfrac{1}{3}\Big)^{10}\Big]}{ \dfrac{2}{3}} \\[1em] = \dfrac{81}{2}\Big[1 - \Big(\dfrac{1}{3}\Big)^{10}\Big] \\[1em] = \dfrac{81}{2}(1 - 3^{-10}). S = ( 1 − r ) a ( 1 − r n ) = 1 − 3 1 27 [ 1 − ( 3 1 ) 10 ] = 3 2 27 [ 1 − ( 3 1 ) 10 ] = 2 81 [ 1 − ( 3 1 ) 10 ] = 2 81 ( 1 − 3 − 10 ) .
Hence, sum = 81 2 ( 1 − 3 − 10 ) . \dfrac{81}{2}(1 - 3^{-10}). 2 81 ( 1 − 3 − 10 ) .
Find a G.P. for which the sum of first two terms is -4 and the fifth term is 4 times the third term.
Answer
Given,
⇒ a + a2 = -4
⇒ a + ar = -4
⇒ a(1 + r) = -4 ........(i)
⇒ a5 = 4a3
⇒ ar4 = 4ar2
⇒ r2 = 4
⇒ r = ± 2
Let r = 2.
Substituting value of r in (i) we get,
⇒ a(1 + 2) = -4
⇒ 3a = -4
⇒ a = − 4 3 -\dfrac{4}{3} − 3 4
G.P. = a, ar, ar2 , ........
= − 4 3 , − 8 3 , − 16 3 -\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3} − 3 4 , − 3 8 , − 3 16
Let r = -2.
Substituting value of r in (i) we get,
⇒ a(1 + (-2)) = -4
⇒ -a = -4
⇒ a = 4.
G.P. = a, ar, ar2 , ........
= 4, -8, 16, -32,........
Hence, G.P. = − 4 3 , − 8 3 , − 16 3 -\dfrac{4}{3}, -\dfrac{8}{3}, -\dfrac{16}{3} − 3 4 , − 3 8 , − 3 16
-1, k and -1 are three consecutive terms of a G.P., then
(i) k = 1
(ii) k = -1
Which of the following is valid ?
only 1
only 2
both 1 and 2
either 1 or -1
Answer
Given,
-1, k and -1 are three consecutive terms of a G.P.
∴ k − 1 = − 1 k ⇒ k × k = − 1 × − 1 ⇒ k 2 = 1 ⇒ k = 1 = ± 1. \therefore \dfrac{k}{-1} = \dfrac{-1}{k} \\[1em] \Rightarrow k \times k = -1 \times -1 \\[1em] \Rightarrow k^2 = 1 \\[1em] \Rightarrow k = \sqrt{1} = \pm 1. ∴ − 1 k = k − 1 ⇒ k × k = − 1 × − 1 ⇒ k 2 = 1 ⇒ k = 1 = ± 1.
Hence, Option 4 is the correct option.
x + 9, 10 and 4 are in G.P. The value of x is :
16
8
-16
0
Answer
Given,
x + 9, 10 and 4 are in G.P.
∴ 10 x + 9 = 4 10 ⇒ 4 ( x + 9 ) = 10 × 10 ⇒ 4 x + 36 = 100 ⇒ 4 x = 100 − 36 ⇒ 4 x = 64 ⇒ x = 64 4 = 16. \therefore \dfrac{10}{x + 9} = \dfrac{4}{10} \\[1em] \Rightarrow 4(x + 9) = 10 \times 10 \\[1em] \Rightarrow 4x + 36 = 100 \\[1em] \Rightarrow 4x = 100 - 36 \\[1em] \Rightarrow 4x = 64 \\[1em] \Rightarrow x = \dfrac{64}{4} = 16. ∴ x + 9 10 = 10 4 ⇒ 4 ( x + 9 ) = 10 × 10 ⇒ 4 x + 36 = 100 ⇒ 4 x = 100 − 36 ⇒ 4 x = 64 ⇒ x = 4 64 = 16.
Hence, Option 1 is the correct option.
The common ratio of a G.P. is 2 and its 6th term is 48. The first term is :
3 2 \dfrac{3}{2} 2 3
2 3 \dfrac{2}{3} 3 2
1
2
Answer
Given,
Common ratio of G.P. (r) = 2
By formula,
⇒ an = arn - 1
Given,
6th term of G.P. is 48.
⇒ a6 = 48
⇒ a × (2)6 - 1 = 48
⇒ a × 25 = 48
⇒ 32a = 48
⇒ a = 48 32 = 3 2 \dfrac{48}{32} = \dfrac{3}{2} 32 48 = 2 3
Hence, Option 1 is the correct option.
Three terms are in G.P., whose product is 27. The middle term is :
-3
3
-3 and 3
-3 or 3
Answer
Let three terms be a r \dfrac{a}{r} r a
Given,
Product of three terms = 27
∴ a r × a × a r = 27 ⇒ a 3 = 3 3 ⇒ a = 3. \therefore \dfrac{a}{r} \times a \times ar = 27 \\[1em] \Rightarrow a^3 = 3^3 \\[1em] \Rightarrow a = 3. ∴ r a × a × a r = 27 ⇒ a 3 = 3 3 ⇒ a = 3.
Hence, Option 2 is the correct option.
The common ratio of a G.P., whose 4th term is 27 and 6th term is 243; is :
9
3
1 3 \dfrac{1}{3} 3 1
1 9 \dfrac{1}{9} 9 1
Answer
Let first term of G.P. be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
4th term of G.P. is 27.
⇒ a4 = 27
⇒ ar4 - 1 = 27
⇒ ar3 = 27 ........(1)
Given,
6th term of G.P. is 243.
⇒ a6 = 243
⇒ ar6 - 1 = 243
⇒ ar5 = 243 ........(2)
Dividing equation (2) by (1), we get :
⇒ a r 5 a r 3 = 243 27 ⇒ r 2 = 9 ⇒ r = 9 = 3. \Rightarrow \dfrac{ar^5}{ar^3} = \dfrac{243}{27} \\[1em] \Rightarrow r^2 = 9 \\[1em] \Rightarrow r = \sqrt{9} = 3. ⇒ a r 3 a r 5 = 27 243 ⇒ r 2 = 9 ⇒ r = 9 = 3.
Hence, Option 2 is the correct option.
The third term of a G.P. = 18, the product of its first five terms is :
18
185
9
18 \sqrt{18} 18
Answer
Let first five terms of G.P. be
a r 2 , a r , a , a r , a r 2 \dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2 r 2 a , r a , a , a r , a r 2
Given,
Third term of a G.P. = 18.
∴ a = 18
Product of first five terms = a r 2 × a r × a × a r × a r 2 \dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2 r 2 a × r a × a × a r × a r 2
= a5
= 185 .
Hence, Option 2 is the correct option.
G.P. : 2 9 , 1 3 , 1 2 , . . . . . . . . . . . . . . \dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2},.............. 9 2 , 3 1 , 2 1 , ..............
Assertion (A): 5th term of the given G.P. is 1 1 8 1\dfrac{1}{8} 1 8 1
Reason (R): If for a G.P., the first term is a, the common ratio is r and the number of terms = n, then sum of the first n terms Sn = a ( r n − 1 ) r − 1 \dfrac{a(r^n - 1)}{r - 1} r − 1 a ( r n − 1 )
A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.
Answer
Given, the sequence = 2 9 , 1 3 , 1 2 , . . . . . . . . . . . . . . \dfrac{2}{9}, \dfrac{1}{3}, \dfrac{1}{2},.............. 9 2 , 3 1 , 2 1 , ..............
First term (a) = 2 9 \dfrac{2}{9} 9 2
Common ratio (r) = 1 3 2 9 = 1 × 9 2 × 3 = 3 2 \dfrac{\dfrac{1}{3}}{\dfrac{2}{9}} = \dfrac{1 \times 9}{2 \times 3} = \dfrac{3}{2} 9 2 3 1 = 2 × 3 1 × 9 = 2 3
Using the formula; Tn = a.rn - 1
T 5 = 2 9 × ( 3 2 ) 5 − 1 = 2 9 × ( 3 2 ) 4 = 2 9 × 81 16 = 9 8 = 1 1 8 . T_5 = \dfrac{2}{9} × \Big(\dfrac{3}{2}\Big)^{5 - 1}\\[1em] = \dfrac{2}{9} × \Big(\dfrac{3}{2}\Big)^4\\[1em] = \dfrac{2}{9} × \dfrac{81}{16}\\[1em] = \dfrac{9}{8}\\[1em] = 1\dfrac{1}{8}. T 5 = 9 2 × ( 2 3 ) 5 − 1 = 9 2 × ( 2 3 ) 4 = 9 2 × 16 81 = 8 9 = 1 8 1 .
So, assertion (A) is true.
According to reason :
Sum of first n terms (Sn ) = a ( r n − 1 ) r − 1 \dfrac{a(r^n - 1)}{r - 1} r − 1 a ( r n − 1 )
But this is not the case when r = 1.
Thus,
Reason (R) is false.
Hence, option 1 is the correct option.
For a G.P., its fourth term = x, seventh term = y and tenth term = z.
Assertion (A): x, y and z are in G.P.
Reason (R): y2 = (ar6 )2 = ar3 × ar9 = xz.
A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.
Answer
Let first term of the G.P. be a and common ratio be r.
By formula :
⇒ Tn = a.rn - 1
Given, fourth term = x, seventh term = y and tenth term = z
⇒ a4 = x, a7 = y and a10 = z
⇒ ar4 - 1 = x, ar7 - 1 = y and ar10 - 1 = z
⇒ ar3 = x, ar6 = y and ar9 = z
If x, y and z are in G.P., then the ratio between the consecutive terms will be equal.
Ratio between y and x :
⇒ y x = a r 6 a r 3 = r 3 . \Rightarrow \dfrac{y}{x} = \dfrac{ar^6}{ar^3} = r^3. ⇒ x y = a r 3 a r 6 = r 3 .
Ratio between z and y :
⇒ z y = a r 9 a r 6 = r 3 . \Rightarrow \dfrac{z}{y} = \dfrac{ar^9}{ar^6} = r^3. ⇒ y z = a r 6 a r 9 = r 3 .
Since, the ratio between the consecutive terms are equal.
Thus, x, y and z are in G.P.
∴ Assertion (A) is true.
⇒ y2 = (ar6 )2
⇒ y2 = ar12
⇒ y2 = ar3 × ar9
⇒ y2 = xz.
∴ Reason (R) is true.
Hence, option 3 is the correct option.
G.P. : = 3 - 6 + 12 - 24 + ............. - 384
Statement (1): Product of 5th term from the beginning and 5th term from the end of the G.P. is -1152.
Statement (2): In an G.P. the product of nth term from the beginning and nth term from the end is
1st term + last term
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Answer
Given, G.P.: = 3 - 6 + 12 - 24 + ............. - 384
Here, a = 3
common ratio, r = − 6 3 \dfrac{-6}{3} 3 − 6
an = -384
Using the formula; Tn = a.rn - 1
⇒ 3 x (-2)n - 1 = -384
⇒ (-2)n - 1 = − 384 3 -\dfrac{384}{3} − 3 384
⇒ (-2)n - 1 = -128
⇒ (-2)n - 1 = (-2)7
⇒ n - 1 = 7
⇒ n = 7 + 1 = 8
5th term from the beginning,
⇒ T5 = 3 x (-2)5 - 1
= 3 x (-2)4
= 3 x 16 = 48
5th term from the last = (8 - 5 + 1) = 4th term from the beginning,
⇒ T4 = 3 x (-2)4 - 1
= 3 x (-2)3
= 3 x (-8) = -24
Product of 5th term from the beginning and 5th term from the end = 48 x (-24) = -1152.
So, statement 1 is true.
Let in a G.P.
a be the first term and N be the total number of terms.
nth term from the beginning,
⇒ Tn = a.rn - 1 ........(1)
nth term from the end,
⇒ TN - n + 1 = a.rN - n + 1 - 1
⇒ TN - n + 1 = arN - n .....(2)
Multiplying equation (1) and (2), we get :
⇒ Tn x TN - n + 1 = a.rn - 1 x a.rN - n
= a2 .r(n - 1) + (N - n)
= a2 .rN - 1 ....................(3)
Now, first term + last term = a + a.rn - 1
= a(1 + rn - 1 ) ....................(4)
From equation (3) and (4),
The product of nth term from the beginning and nth term from the end is not equal to 1st term + last term.
So, statement 2 is false.
Hence, option 3 is the correct option.
In a G.P., common ratio = 2, first term = 3 and last term = 96.
Statement (1): The number of terms in this G.P. = 96 - 3.
Statement (2): a : arn - 1 = 3 : 96
Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.
Answer
Given,
First term, a = 3
Common ratio, r = 2
Last term = 96
Using the formula;
⇒ Tn = a.rn - 1
⇒ 3 x 2n - 1 = 96
⇒ 2n - 1 = 96 3 \dfrac{96}{3} 3 96
⇒ 2n - 1 = 32
⇒ 2n - 1 = 25
⇒ n - 1 = 5
⇒ n = 5 + 1 = 6
And, according to statement 1, the number of terms in this G.P. = 96 - 3 = 93, which is not correct.
So, statement 1 is false.
⇒ a a r n − 1 = first term last term = 3 96 = 3 : 96. \Rightarrow \dfrac{a}{ar^{n - 1}} = \dfrac{\text{first term}}{\text{last term}}\\[1em] = \dfrac{3}{96} = 3 : 96. ⇒ a r n − 1 a = last term first term = 96 3 = 3 : 96.
So, statement 2 is true.
Hence, option 4 is the correct option.
The 5th and the 8th terms of a G.P. are 32 and 256 respectively. Find its first term and the common ratio.
Answer
Let first term of G.P. be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
5th term of G.P. = 32
⇒ a5 = 32
⇒ ar5 - 1 = 32
⇒ ar4 = 32 ..........(1)
Given,
8th term of G.P. = 256
⇒ a8 = 256
⇒ ar8 - 1 = 256
⇒ ar7 = 256 ..........(2)
Dividing equation (2) by (1), we get :
⇒ a r 7 a r 4 = 256 32 ⇒ r 3 = 8 ⇒ r 3 = 2 3 ⇒ r = 2. \Rightarrow \dfrac{ar^7}{ar^4} = \dfrac{256}{32} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2. ⇒ a r 4 a r 7 = 32 256 ⇒ r 3 = 8 ⇒ r 3 = 2 3 ⇒ r = 2.
Substituting value of r in equation (1), we get :
⇒ a × (2)4 = 32
⇒ 16a = 32
⇒ a = 32 16 \dfrac{32}{16} 16 32
Hence, first term = 2 and common ratio = 2.
The third term of a G.P. is greater than its first term by 9 whereas its second term is greater than the fourth term by 18. Find the G.P.
Answer
Let first term be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
The third term of a G.P. is greater than its first term by 9.
∴ a3 - a = 9
⇒ ar3 - 1 - a = 9
⇒ ar2 - a = 9
⇒ a(r2 - 1) = 9
⇒ a = 9 r 2 − 1 \dfrac{9}{r^2 - 1} r 2 − 1 9
Given,
Second term is greater than the fourth term by 18.
∴ a2 - a4 = 18
⇒ ar2 - 1 - ar4 - 1 = 18
⇒ ar - ar3 = 18
⇒ ar(1 - r2 ) = 18
Substituting value of a from equation (1) in above equation, we get :
⇒ 9 r 2 − 1 × r ( 1 − r 2 ) = 18 ⇒ 9 r 2 − 1 × − r ( r 2 − 1 ) = 18 ⇒ − 9 r = 18 ⇒ r = − 18 9 = − 2. \Rightarrow \dfrac{9}{r^2 - 1} \times r(1 - r^2) = 18 \\[1em] \Rightarrow \dfrac{9}{r^2 - 1} \times -r(r^2 - 1) = 18 \\[1em] \Rightarrow -9r = 18 \\[1em] \Rightarrow r = -\dfrac{18}{9} = -2. ⇒ r 2 − 1 9 × r ( 1 − r 2 ) = 18 ⇒ r 2 − 1 9 × − r ( r 2 − 1 ) = 18 ⇒ − 9 r = 18 ⇒ r = − 9 18 = − 2.
Substituting value of r in equation (1), we get :
⇒ a = 9 ( − 2 ) 2 − 1 = 9 4 − 1 = 9 3 = 3. \Rightarrow a = \dfrac{9}{(-2)^2 - 1} \\[1em] = \dfrac{9}{4 - 1} \\[1em] = \dfrac{9}{3} \\[1em] = 3. ⇒ a = ( − 2 ) 2 − 1 9 = 4 − 1 9 = 3 9 = 3.
G.P. = a, ar, ar2 , ar3 , .......
= 3, 3 × -2, 3 × (-2)2 , 3 × (-2)3 , .......
= 3, -6, 3 × 4, 3 × -8, ........
= 3, -6, 12, -24, ........
Hence, required G.P. = 3, -6, 12, -24, ........
x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P. Find the value of G.
Answer
Given,
x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P.
∴ 2 x + 2 x = 3 x + 3 2 x + 2 ⇒ ( 2 x + 2 ) 2 = x ( 3 x + 3 ) ⇒ ( 2 x ) 2 + 2 2 + 2 × 2 x × 2 = 3 x 2 + 3 x ⇒ 4 x 2 + 4 + 8 x = 3 x 2 + 3 x ⇒ 4 x 2 − 3 x 2 + 8 x − 3 x + 4 = 0 ⇒ x 2 + 5 x + 4 = 0 ⇒ x 2 + 4 x + x + 4 = 0 ⇒ x ( x + 4 ) + 1 ( x + 4 ) = 0 ⇒ ( x + 1 ) ( x + 4 ) = 0 ⇒ x + 1 = 0 or x + 4 = 0 ⇒ x = − 1 or x = − 4. \therefore \dfrac{2x + 2}{x} = \dfrac{3x + 3}{2x + 2} \\[1em] \Rightarrow (2x + 2)^2 = x(3x + 3) \\[1em] \Rightarrow (2x)^2 + 2^2 + 2 \times 2x \times 2 = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 + 4 + 8x = 3x^2 + 3x \\[1em] \Rightarrow 4x^2 - 3x^2 + 8x - 3x + 4 = 0 \\[1em] \Rightarrow x^2 + 5x + 4 = 0 \\[1em] \Rightarrow x^2 + 4x + x + 4 = 0 \\[1em] \Rightarrow x(x + 4) + 1(x + 4) = 0 \\[1em] \Rightarrow (x + 1)(x + 4) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x + 4 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = -4. ∴ x 2 x + 2 = 2 x + 2 3 x + 3 ⇒ ( 2 x + 2 ) 2 = x ( 3 x + 3 ) ⇒ ( 2 x ) 2 + 2 2 + 2 × 2 x × 2 = 3 x 2 + 3 x ⇒ 4 x 2 + 4 + 8 x = 3 x 2 + 3 x ⇒ 4 x 2 − 3 x 2 + 8 x − 3 x + 4 = 0 ⇒ x 2 + 5 x + 4 = 0 ⇒ x 2 + 4 x + x + 4 = 0 ⇒ x ( x + 4 ) + 1 ( x + 4 ) = 0 ⇒ ( x + 1 ) ( x + 4 ) = 0 ⇒ x + 1 = 0 or x + 4 = 0 ⇒ x = − 1 or x = − 4.
Substituting value of x = -1, in terms we get :
Terms : -1, 2(-1) + 2, 3(-1) + 3, G
= -1, -2 + 2, -3 + 3, G
= -1, 0, 0, G
This is not possible as in this case common ratio is different.
Substituting value of x = -4, in terms we get :
Terms : -4, 2(-4) + 2, 3(-4) + 3, G
= -4, -8 + 2, -12 + 3, G
= -4, -6, -9, G
Here, common difference = − 6 − 4 = 3 2 \dfrac{-6}{-4} = \dfrac{3}{2} − 4 − 6 = 2 3
G = − 9 × 3 2 = − 27 2 -9 \times \dfrac{3}{2} = -\dfrac{27}{2} − 9 × 2 3 = − 2 27
Hence, G = − 27 2 -\dfrac{27}{2} − 2 27
The third term of a G.P. is 2. Find the product of the first five terms of this G.P.
Answer
Let first five terms of G.P. be a r 2 , a r , a , a r , a r 2 \dfrac{a}{r^2}, \dfrac{a}{r}, a, ar, ar^2 r 2 a , r a , a , a r , a r 2
Given,
Third term of the G.P. is 2.
∴ a = 2
Product of terms = a r 2 × a r × a × a r × a r 2 = a 5 \dfrac{a}{r^2} \times \dfrac{a}{r} \times a \times ar \times ar^2 = a^5 r 2 a × r a × a × a r × a r 2 = a 5
= 25
= 32.
Hence, product of the first five terms of this G.P. = 32.
The 10th , 16th and 22nd terms of a G.P. are x, y and z respectively. Show that x, y and z are in G.P.
Answer
Let first term of G.P. be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
The 10th , 16th and 22nd terms of a G.P. are x, y and z respectively.
⇒ a10 = x
⇒ x = ar10 - 1
⇒ x = ar9 ...........(1)
⇒ a16 = y
⇒ y = ar16 - 1
⇒ y = ar15 ...........(2)
⇒ a22 = z
⇒ z = ar22 - 1
⇒ z = ar21 ...........(3)
Dividing equation (2) by (1), we get :
⇒ y x = a r 15 a r 9 ⇒ y x = r 15 r 9 ⇒ y x = r 15 − 9 ⇒ y x = r 6 . \Rightarrow \dfrac{y}{x} = \dfrac{ar^{15}}{ar^9} \\[1em] \Rightarrow \dfrac{y}{x} = \dfrac{r^{15}}{r^9} \\[1em] \Rightarrow \dfrac{y}{x} = r^{15 - 9} \\[1em] \Rightarrow \dfrac{y}{x} = r^6. ⇒ x y = a r 9 a r 15 ⇒ x y = r 9 r 15 ⇒ x y = r 15 − 9 ⇒ x y = r 6 .
Dividing equation (3) by (2), we get :
⇒ z y = a r 21 a r 15 ⇒ z y = r 21 r 15 ⇒ z y = r 21 − 15 ⇒ z y = r 6 . \Rightarrow \dfrac{z}{y} = \dfrac{ar^{21}}{ar^{15}} \\[1em] \Rightarrow \dfrac{z}{y} = \dfrac{r^{21}}{r^{15}} \\[1em] \Rightarrow \dfrac{z}{y} = r^{21 - 15} \\[1em] \Rightarrow \dfrac{z}{y} = r^6. ⇒ y z = a r 15 a r 21 ⇒ y z = r 15 r 21 ⇒ y z = r 21 − 15 ⇒ y z = r 6 .
Since, y x = z y \dfrac{y}{x} = \dfrac{z}{y} x y = y z 6 .
Hence, proved that x, y and z are in G.P.
Which term of the G.P. 2 , 2 2 , 4 , . . . . . . . . is 128 2 ? 2, 2\sqrt{2}, 4, ........ \text{ is } 128\sqrt{2} ? 2 , 2 2 , 4 , ........ is 128 2 ?
Answer
Given,
G.P. : 2 , 2 2 , 4 , . . . . . . . . 2, 2\sqrt{2}, 4, ........ 2 , 2 2 , 4 , ........
In above G.P.,
First term (a) = 2
Common ratio (r) = 2 2 2 = 2 \dfrac{2\sqrt{2}}{2} = \sqrt{2} 2 2 2 = 2
Let nth term of G.P. be 128 2 128\sqrt{2} 128 2
⇒ a r n − 1 = 128 2 ⇒ 2 × ( 2 ) n − 1 = 128 2 ⇒ 2 × ( 2 1 2 ) n − 1 = ( 2 ) 7 . 2 ⇒ 2 × ( 2 ) n − 1 2 = 2 7 .2 1 2 ⇒ ( 2 ) 1 + n − 1 2 = ( 2 ) 7 + 1 2 ⇒ ( 2 ) n − 1 + 2 2 = ( 2 ) 14 + 1 2 ⇒ ( 2 ) n + 1 2 = ( 2 ) 15 2 ⇒ n + 1 2 = 15 2 ⇒ n + 1 = 15 ⇒ n = 15 − 1 = 14. \Rightarrow ar^{n-1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (\sqrt{2})^{n - 1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (2^{\dfrac{1}{2}})^{n - 1} = (2)^7.\sqrt{2} \\[1em] \Rightarrow 2 \times (2)^{\dfrac{n - 1}{2}} = 2^7.2^{\dfrac{1}{2}} \\[1em] \Rightarrow (2)^{1 + \dfrac{n - 1}{2}} = (2)^{7 + \dfrac{1}{2}} \\[1em] \Rightarrow (2)^{\dfrac{n - 1 + 2}{2}} = (2)^{\dfrac{14 + 1}{2}} \\[1em] \Rightarrow (2)^{\dfrac{n + 1}{2}} = (2)^{\dfrac{15}{2}} \\[1em] \Rightarrow \dfrac{n + 1}{2} = \dfrac{15}{2} \\[1em] \Rightarrow n + 1 = 15 \\[1em] \Rightarrow n = 15 - 1 = 14. ⇒ a r n − 1 = 128 2 ⇒ 2 × ( 2 ) n − 1 = 128 2 ⇒ 2 × ( 2 2 1 ) n − 1 = ( 2 ) 7 . 2 ⇒ 2 × ( 2 ) 2 n − 1 = 2 7 . 2 2 1 ⇒ ( 2 ) 1 + 2 n − 1 = ( 2 ) 7 + 2 1 ⇒ ( 2 ) 2 n − 1 + 2 = ( 2 ) 2 14 + 1 ⇒ ( 2 ) 2 n + 1 = ( 2 ) 2 15 ⇒ 2 n + 1 = 2 15 ⇒ n + 1 = 15 ⇒ n = 15 − 1 = 14.
Hence, 14th term of G.P. = 128 2 128\sqrt{2} 128 2
Find the 8th term of a G.P., if its common ratio is 2 and 10th term is 768.
Answer
Given,
Common ratio (r) = 2
By formula,
⇒ an = arn - 1
Given,
10th term of G.P. = 768
⇒ a10 = 768
⇒ ar10 - 1 = 768
⇒ ar9 = 768
⇒ a × 29 = 768
⇒ a × 512 = 768
⇒ a = 768 512 = 3 2 \dfrac{768}{512} = \dfrac{3}{2} 512 768 = 2 3
8th term of G.P. is given by :
⇒ a8 = ar8 - 1
⇒ a8 = ar7
⇒ a8 = 3 2 × 2 7 \dfrac{3}{2} \times 2^7 2 3 × 2 7
⇒ a8 = 3 × 26
⇒ a8 = 3 × 64 = 192.
Hence, 8th term of a G.P. is given by 192.
In a G.P., the 4th term is 48 and 7th term is 384. Find its 6th term.
Answer
Let first term of G.P. be a and common ratio be r.
By formula,
⇒ an = arn - 1
Given,
4th term of G.P. = 48
⇒ a4 = 48
⇒ ar4 - 1 = 48
⇒ ar3 = 48 ..........(1)
Given,
7th term of G.P. = 384
⇒ a7 = 48
⇒ ar7 - 1 = 384
⇒ ar6 = 384 ..........(2)
Dividing equation (2) by (1), we get :
⇒ a r 6 a r 3 = 384 48 ⇒ r 3 = 8 ⇒ r 3 = 2 3 ⇒ r = 2. \Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{384}{48} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = 2^3 \\[1em] \Rightarrow r = 2. ⇒ a r 3 a r 6 = 48 384 ⇒ r 3 = 8 ⇒ r 3 = 2 3 ⇒ r = 2.
Substituting value of r in equation (1), we get :
⇒ a(2)3 = 48
⇒ a × 8 = 48
⇒ a = 48 8 \dfrac{48}{8} 8 48
⇒ a6 = ar6 - 1
⇒ a6 = ar5
⇒ a6 = 6 × 25
⇒ a6 = 6 × 32
⇒ a6 = 192.
Hence, 6th term of G.P. = 192.
− 5 3 , x and − 3 5 -\dfrac{5}{3}, x \text{ and } -\dfrac{3}{5} − 3 5 , x and − 5 3
Answer
Given,
− 5 3 , x and − 3 5 -\dfrac{5}{3}, x \text{ and } -\dfrac{3}{5} − 3 5 , x and − 5 3
∴ x − 5 3 = − 3 5 x ⇒ x 2 = − 3 5 × − 5 3 ⇒ x 2 = 1 ⇒ x = 1 = ± 1. \therefore \dfrac{x}{-\dfrac{5}{3}} = \dfrac{-\dfrac{3}{5}}{x} \\[1em] \Rightarrow x^2= -\dfrac{3}{5} \times -\dfrac{5}{3} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x = \sqrt{1} = \pm 1. ∴ − 3 5 x = x − 5 3 ⇒ x 2 = − 5 3 × − 3 5 ⇒ x 2 = 1 ⇒ x = 1 = ± 1.
Hence, x = +1 or -1.
Rohit writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with the instructions that they move the chain similarly. Assuming that the chain is not broken and that it costs ₹ 10 to mail one letter, determine the amount spent on postage when the fourth set of letters is mailed.
Answer
Rohit writes a letter to four of his friends, and each of them copies the letter and mail it to four different persons and the chain continues. Thus, the above sequence is :
4, 16, 64, .......
The above sequence is a G.P., with first term (a) = 4 and r = 4.
By formula,
Sum of G.P. = a ( r n − 1 ) ( r − 1 ) \dfrac{a(r^n - 1)}{(r - 1)} ( r − 1 ) a ( r n − 1 )
Thus,
S 4 = 4 × ( 4 4 − 1 ) 4 − 1 = 4 × ( 256 − 1 ) 3 = 4 × 255 3 = 4 × 85 = 340. S_4 = \dfrac{4 \times (4^4 - 1)}{4 - 1} \\[1em] = \dfrac{4 \times (256 - 1)}{3} \\[1em] = \dfrac{4 \times 255}{3} \\[1em] = 4 \times 85 \\[1em] = 340. S 4 = 4 − 1 4 × ( 4 4 − 1 ) = 3 4 × ( 256 − 1 ) = 3 4 × 255 = 4 × 85 = 340.
Thus, total 340 letters are mailed till fourth set.
If, it costs ₹ 10 to mail one letter.
Total cost = No. of letters × 10 = 340 × ₹ 10 = ₹ 3,400.
Hence, amount spent on postage = ₹ 3,400.
The fourth term of a G.P. is eight times its seventh term. The fifth term of then G.P. is 3 16 \dfrac{3}{16} 16 3 th term.
Answer
By formula,
an = arn - 1
Given,
The fourth term of a G.P. is eight times its seventh term.
a4 = 8a7
ar4 - 1 = 8ar7 - 1
ar3 = 8ar6
r 6 r 3 = a 8 a \dfrac{r^6}{r^3} = \dfrac{a}{8a} r 3 r 6 = 8 a a
r 3 = 1 8 r^3 = \dfrac{1}{8} r 3 = 8 1
r = 1 8 3 = 1 2 \sqrt[3]{\dfrac{1}{8}} = \dfrac{1}{2} 3 8 1 = 2 1
Given,
Fifth term = 3 16 \dfrac{3}{16} 16 3
⇒ a r 4 = 3 16 ⇒ a × ( 1 2 ) 4 = 3 16 ⇒ a × 1 16 = 3 16 ⇒ a = 3. \Rightarrow ar^4 = \dfrac{3}{16} \\[1em] \Rightarrow a \times \Big(\dfrac{1}{2}\Big)^4 = \dfrac{3}{16} \\[1em] \Rightarrow a \times \dfrac{1}{16} = \dfrac{3}{16} \\[1em] \Rightarrow a = 3. ⇒ a r 4 = 16 3 ⇒ a × ( 2 1 ) 4 = 16 3 ⇒ a × 16 1 = 16 3 ⇒ a = 3.
12th term :
a12 = ar12 - 1
= ar11
= 3 × ( 1 2 ) 11 3 \times \Big(\dfrac{1}{2}\Big)^{11} 3 × ( 2 1 ) 11
Hence, 12th term = 3 × ( 1 2 ) 11 3 \times \Big(\dfrac{1}{2}\Big)^{11} 3 × ( 2 1 ) 11
Richard borrows ₹ 10,000 from his friend Ramesh. Ramesh asks him to return the money just by repaying ₹ 10 at first day and double the amount in the subsequent days than the each previous day continuously for 10 days. Richard thanks Ramesh for his help. Determine whether Ramesh was in loss or profit.
Answer
Sequence :
10 + 20 + 40 + ......... upto 10 terms.
The above is a G.P. with first term (a) = 10 and common ratio (r) = 2.
By formula,
Sn = a ( r n − 1 ) ( r − 1 ) \dfrac{a(r^n - 1)}{(r - 1)} ( r − 1 ) a ( r n − 1 )
Substituting values we get :
⇒ S 10 = 10 × ( 2 10 − 1 ) 2 − 1 = 10 × ( 1024 − 1 ) = 10 × 1023 = 10 , 230. \Rightarrow S_{10} = \dfrac{10 \times (2^{10} - 1)}{2 - 1} \\[1em] = 10 \times (1024 - 1) \\[1em] = 10 \times 1023 \\[1em] = 10,230. ⇒ S 10 = 2 − 1 10 × ( 2 10 − 1 ) = 10 × ( 1024 − 1 ) = 10 × 1023 = 10 , 230.
Since, Ramesh lent ₹ 10,000 but received ₹ 10,230 thus he is in profit of ₹ 230.
Hence, Ramesh makes a profit of ₹ 230.
Case Study Based Question :
Case study:
Based on the above information, answer the following questions:
(i) The distance covered by Rohan in placing the flags on one side of the middle most flag forms a sequence. Identify the type of sequence by writing at least the first 4 terms of the sequence.
(ii) Find the distance covered by Rohan in placing the 10th flag and coming back to the middle most flag.
(iii) Find the total distance travelled by Rohan in fixing all the 27 flags.
Answer
(i) There are 27 flags in total. The middle-most flag is the 14th flag.
This means there are 13 flags to the left and 13 flags to the right of the starting point.
The first flag on either side is 3 meters away. Rohan travels 3 m to place it and 3 m to return.
The second flag is 6 m away. Rohan travels 6 m to place it and 6 m to return.
The first 4 terms (round-trip distances) are :
1st flag = 2 × 3 = 6 m
2nd flag = 2 × 6 = 12 m
3rd flag = 2 × 9 = 18 m
4th flag = 2 × 12 = 24 m
The above terms form an A.P. with common difference = 6.
Hence, the required sequence is an A.P.
(ii) By formula,
an = a + (n - 1)d
a10 = 6 + 6(10 - 1)
= 6 + 6 × 9
= 6 + 54
= 60 m.
Hence, the distance covered by Rohan in placing the 10th flag and returning back to the middle most flag = 60 m.
(iii) By formula,
Sn = n 2 [ 2 a + ( n − 1 ) d ] \dfrac{n}{2}[2a + (n - 1)d] 2 n [ 2 a + ( n − 1 ) d ]
On each side 13 flags are placed.
S 13 = 13 2 [ 2 × 6 + ( 13 − 1 ) × 6 ] = 13 2 × [ 12 + 12 × 6 ] = 13 2 × [ 12 + 72 ] = 13 2 × 84 = 13 × 42 = 546 m . S_{13} = \dfrac{13}{2}[2 \times 6 + (13 - 1) \times 6] \\[1em] = \dfrac{13}{2} \times [12 + 12 \times 6] \\[1em] = \dfrac{13}{2} \times [12 + 72] \\[1em] = \dfrac{13}{2} \times 84 \\[1em] = 13 \times 42 \\[1em] = 546 \text{ m}. S 13 = 2 13 [ 2 × 6 + ( 13 − 1 ) × 6 ] = 2 13 × [ 12 + 12 × 6 ] = 2 13 × [ 12 + 72 ] = 2 13 × 84 = 13 × 42 = 546 m .
Thus, distance covered on one side = 546 m.
Total distance = 2 × 546 = 1092 m.
Hence, total distance travelled by Rohan in fixing all the 27 flags = 1092 m.
