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.
The geometric mean between 8 and 32 is :
24
256
40
16
Answer
By formula,
Geometric mean between a and b = a b \sqrt{ab} ab
∴ Geometric mean between 8 and 32 = 8 × 32 = 256 \sqrt{8 \times 32} = \sqrt{256} 8 × 32 = 256
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.
Find the geometric mean between
(i) 4 9 and 9 4 \dfrac{4}{9} \text{ and } \dfrac{9}{4} 9 4 and 4 9
(ii) 14 and 7 32 \dfrac{7}{32} 32 7
(iii) 2a and 8a3
Answer
(i) Geometric mean between 4 9 and 9 4 \dfrac{4}{9} \text{ and } \dfrac{9}{4} 9 4 and 4 9
G = 4 9 × 9 4 = 1 = 1. G = \sqrt{\dfrac{4}{9} \times \dfrac{9}{4}} \\[1em] = \sqrt{1} \\[1em] = 1. G = 9 4 × 4 9 = 1 = 1.
Hence, geometric mean = 1.
(ii) Geometric mean between 14 and 7 32 \dfrac{7}{32} 32 7
G = 14 × 7 32 = 49 16 = 7 4 . G = \sqrt{14 \times \dfrac{7}{32}} \\[1em] = \sqrt{\dfrac{49}{16}} \\[1em] = \dfrac{7}{4}. G = 14 × 32 7 = 16 49 = 4 7 .
Hence, geometric mean = 7 4 = 1 3 4 \dfrac{7}{4} = 1\dfrac{3}{4} 4 7 = 1 4 3
(iii) Geometric mean between 2a and 8a3
G = 2 a × 8 a 3 = 16 a 4 = 4 a 2 . G = \sqrt{2a \times 8a^3} \\[1em] = \sqrt{16a^4} \\[1em] = 4a^2. G = 2 a × 8 a 3 = 16 a 4 = 4 a 2 .
Hence, geometric mean = 4a2
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 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 term 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.
When first term = a, common ratio = r then
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 )
So, reason (R) is true, but it doesnot explain assertion.
Hence, option 4 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.
