Find the sum of :
(i) 20 terms of the series 2 + 6 + 18 + ...
(ii) 10 terms of the series 1 + 3 \sqrt{3} 3
(iii) 6 terms of the G.P. 1, − 2 3 , 4 9 , -\dfrac{2}{3}, \dfrac{4}{9}, − 3 2 , 9 4 ,
(iv) 5 terms and n terms of the series 1 + 2 3 + 4 9 + . . . . \dfrac{2}{3} + \dfrac{4}{9} + .... 3 2 + 9 4 + ....
Answer
(i) Sum = 2 + 6 + 18 + ..... + (20th term)
The above list of numbers is a G.P. with first term a = 2 and common ratio = r = 6 2 = 3. \dfrac{6}{2} = 3. 2 6 = 3.
By formula,
S n = a ( r n − 1 ) r − 1 ∴ S 20 = 2 ( ( 3 ) 20 − 1 ) 3 − 1 = 2 ( 3 20 − 1 ) 2 = 3 20 − 1. S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{20} = \dfrac{2((3)^{20} - 1)}{3 - 1} \\[1em] = \dfrac{2(3^{20} - 1)}{2} \\[1em] = 3^{20} - 1. S n = r − 1 a ( r n − 1 ) ∴ S 20 = 3 − 1 2 (( 3 ) 20 − 1 ) = 2 2 ( 3 20 − 1 ) = 3 20 − 1.
Hence, the sum of the series is 320 - 1.
(ii) Sum = 1 + 3 \sqrt{3} 3
The above list of numbers is a G.P. with first term a = 1 and common ratio = r = 3 . \sqrt{3}. 3 .
By formula,
S n = a ( r n − 1 ) r − 1 ∴ S 10 = 1 ( ( 3 ) 10 − 1 ) 3 − 1 = ( ( 3 ) 10 / 2 − 1 ) 3 − 1 = 3 5 − 1 3 − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{10} = \dfrac{1((\sqrt{3})^{10} - 1)}{\sqrt{3} - 1} \\[1em] = \dfrac{((3)^{10/2} - 1)}{\sqrt{3} - 1} \\[1em] = \dfrac{3^5 - 1}{\sqrt{3} - 1} S n = r − 1 a ( r n − 1 ) ∴ S 10 = 3 − 1 1 (( 3 ) 10 − 1 ) = 3 − 1 (( 3 ) 10/2 − 1 ) = 3 − 1 3 5 − 1
Multiplying numerator and denominator by 3 \sqrt{3} 3
= 3 5 − 1 3 − 1 × 3 + 1 3 + 1 = 242 ( 3 + 1 ) ( 3 − 1 ) ( 3 + 1 ) = 242 ( 3 + 1 ) 3 + 3 − 3 − 1 = 242 ( 3 + 1 ) 2 = 121 ( 3 + 1 ) . = \dfrac{3^5 - 1}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] = \dfrac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] = \dfrac{242(\sqrt{3} + 1)}{3 + \sqrt{3} - \sqrt{3} - 1} \\[1em] = \dfrac{242(\sqrt{3} + 1)}{2} \\[1em] = 121(\sqrt{3} + 1). = 3 − 1 3 5 − 1 × 3 + 1 3 + 1 = ( 3 − 1 ) ( 3 + 1 ) 242 ( 3 + 1 ) = 3 + 3 − 3 − 1 242 ( 3 + 1 ) = 2 242 ( 3 + 1 ) = 121 ( 3 + 1 ) .
Hence, the sum of the series is 121 ( 3 + 1 ) 121(\sqrt{3} + 1) 121 ( 3 + 1 )
(iii) Sum = 1, − 2 3 , 4 9 , -\dfrac{2}{3}, \dfrac{4}{9}, − 3 2 , 9 4 ,
The above list of numbers is a G.P. with first term a = 1 and common ratio = r = − 2 3 . -\dfrac{2}{3}. − 3 2 .
By formula,
S n = a ( r n − 1 ) r − 1 ∴ S 6 = 1 [ ( − 2 3 ) 6 − 1 ] − 2 3 − 1 = 2 6 3 6 − 1 − 2 − 3 3 = 2 6 3 6 − 1 − 5 3 = 3 ( 2 6 − 3 6 ) 3 6 × − 5 = 64 − 729 3 5 × − 5 = − 665 − 1215 S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{6} = \dfrac{1\Big[\Big(-\dfrac{2}{3}\Big)^6 - 1\Big]}{-\dfrac{2}{3} - 1} \\[1em] = \dfrac{\dfrac{2^6}{3^6} - 1}{\dfrac{-2 -3}{3}} \\[1em] = \dfrac{\dfrac{2^6}{3^6} - 1}{-\dfrac{5}{3}} \\[1em] = \dfrac{3(2^6 - 3^6)}{3^6 \times -5} \\[1em] = \dfrac{64 - 729}{3^5 \times -5} \\[1em] = \dfrac{-665}{-1215} S n = r − 1 a ( r n − 1 ) ∴ S 6 = − 3 2 − 1 1 [ ( − 3 2 ) 6 − 1 ] = 3 − 2 − 3 3 6 2 6 − 1 = − 3 5 3 6 2 6 − 1 = 3 6 × − 5 3 ( 2 6 − 3 6 ) = 3 5 × − 5 64 − 729 = − 1215 − 665
On dividing numerator and denominator by -5 we get,
= 133 243 . = \dfrac{133}{243}. = 243 133 .
Hence, the sum of the series is 133 243 \dfrac{133}{243} 243 133
(iv) Sum upto 5 terms = 1 + 2 3 + 4 9 + . . . . \dfrac{2}{3} + \dfrac{4}{9} + .... 3 2 + 9 4 + ....
The above list of numbers is a G.P. with first term a = 1 and common ratio = r = 2 3 . \dfrac{2}{3}. 3 2 .
By formula,
S n = a ( r n − 1 ) r − 1 ∴ S 5 = 1 [ ( 2 3 ) 5 − 1 ] 2 3 − 1 = 2 5 3 5 − 1 2 − 3 3 = 2 5 3 5 − 1 − 1 3 = 3 ( 2 5 − 3 5 ) 3 5 × − 1 = 32 − 243 3 4 × − 1 = − 211 − 81 S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_5 = \dfrac{1\Big[\Big(\dfrac{2}{3}\Big)^5 - 1\Big]}{\dfrac{2}{3} - 1} \\[1em] = \dfrac{\dfrac{2^5}{3^5} - 1}{\dfrac{2 -3}{3}} \\[1em] = \dfrac{\dfrac{2^5}{3^5} - 1}{-\dfrac{1}{3}} \\[1em] = \dfrac{3(2^5 - 3^5)}{3^5 \times -1} \\[1em] = \dfrac{32 - 243}{3^4 \times -1} \\[1em] = \dfrac{-211}{-81} S n = r − 1 a ( r n − 1 ) ∴ S 5 = 3 2 − 1 1 [ ( 3 2 ) 5 − 1 ] = 3 2 − 3 3 5 2 5 − 1 = − 3 1 3 5 2 5 − 1 = 3 5 × − 1 3 ( 2 5 − 3 5 ) = 3 4 × − 1 32 − 243 = − 81 − 211
On dividing numerator and denominator by -1 we get,
= 211 81 = \dfrac{211}{81} = 81 211
Sum of n terms = S n = 1 [ ( 2 3 ) n − 1 ] 2 3 − 1 S_n = \dfrac{1\Big[\Big(\dfrac{2}{3}\Big)^n - 1\Big]}{\dfrac{2}{3} - 1} S n = 3 2 − 1 1 [ ( 3 2 ) n − 1 ]
= ( 2 3 ) n − 1 2 − 3 3 = 3 [ ( 2 3 ) n − 1 ] − 1 = − 3 [ ( 2 3 ) n − 1 ] = 3 [ 1 − ( 2 3 ) n ] . = \dfrac{\Big(\dfrac{2}{3}\Big)^n - 1}{\dfrac{2 - 3}{3}} \\[1em] = \dfrac{3\Big[\Big(\dfrac{2}{3}\Big)^n - 1\Big]}{-1} \\[1em] = -3\Big[\Big(\dfrac{2}{3}\Big)^n - 1\Big] \\[1em] = 3\Big[1 - \Big(\dfrac{2}{3}\Big)^n\Big]. = 3 2 − 3 ( 3 2 ) n − 1 = − 1 3 [ ( 3 2 ) n − 1 ] = − 3 [ ( 3 2 ) n − 1 ] = 3 [ 1 − ( 3 2 ) n ] .
Hence, the sum of the series upto 5 terms is 211 81 \dfrac{211}{81} 81 211 3 [ 1 − ( 2 3 ) n ] 3\Big[1 - \Big(\dfrac{2}{3}\Big)^n\Big] 3 [ 1 − ( 3 2 ) n ]
Find the sum of the series 81 - 27 + 9 - ..... - 1 27 \dfrac{1}{27} 27 1
Answer
The above series is in G.P. with a = 81, r = − 1 3 and l = − 1 27 . -\dfrac{1}{3} \text{ and l } = -\dfrac{1}{27}. − 3 1 and l = − 27 1 .
Sum = a − l r 1 − r ∴ Sum = 81 − ( − 1 27 ) ( − 1 3 ) 1 − ( − 1 3 ) = 81 − 1 81 1 + 1 3 = 6561 − 1 81 3 + 1 3 = 6560 81 4 3 = 6560 × 3 81 × 4 = 19680 324 \text{Sum } = \dfrac{a - lr}{1 - r} \\[1em] \therefore \text{Sum } = \dfrac{81 - \Big(-\dfrac{1}{27}\Big)(-\dfrac{1}{3})}{1 - \Big(-\dfrac{1}{3}\Big)} \\[1em] = \dfrac{81 - \dfrac{1}{81}}{1 + \dfrac{1}{3}} \\[1em] = \dfrac{\dfrac{6561 - 1}{81}}{\dfrac{3 + 1}{3}} \\[1em] = \dfrac{\dfrac{6560}{81}}{\dfrac{4}{3}} \\[1em] = \dfrac{6560 \times 3}{81 \times 4} \\[1em] = \dfrac{19680}{324} Sum = 1 − r a − l r ∴ Sum = 1 − ( − 3 1 ) 81 − ( − 27 1 ) ( − 3 1 ) = 1 + 3 1 81 − 81 1 = 3 3 + 1 81 6561 − 1 = 3 4 81 6560 = 81 × 4 6560 × 3 = 324 19680
Dividing numerator and denominator by 12 we get,
= 1640 27 . = \dfrac{1640}{27}. = 27 1640 .
Hence, the sum of the series 81 - 27 + 9 - ..... - 1 27 is 1640 27 . \dfrac{1}{27} \text{ is } \dfrac{1640}{27}. 27 1 is 27 1640 .
The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.
Answer
Given, an = 128, r = 2 and Sn = 255.
By formula an = arn - 1 n - 1
⇒ a 2 n 2 = 128 ⇒ 2 n = 256 a . (Eq 1) \Rightarrow a\dfrac{2^n}{2} = 128 \\[1em] \Rightarrow 2^n = \dfrac{256}{a}. \text{ (Eq 1) } ⇒ a 2 2 n = 128 ⇒ 2 n = a 256 . (Eq 1)
By formula,
S n = a ( r n − 1 ) r − 1 ∴ 255 = a ( 2 n − 1 ) 2 − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore 255 = \dfrac{a(2^n - 1)}{2 - 1} S n = r − 1 a ( r n − 1 ) ∴ 255 = 2 − 1 a ( 2 n − 1 )
Putting value of 2n from Eq 1 in above Equation,
⇒ 255 = a ( 256 a − 1 ) ⇒ 255 = 256 − a ⇒ a = 256 − 255 = 1. \Rightarrow 255 = a\Big(\dfrac{256}{a} - 1\Big) \\[1em] \Rightarrow 255 = 256 - a \\[1em] \Rightarrow a = 256 - 255 = 1. ⇒ 255 = a ( a 256 − 1 ) ⇒ 255 = 256 − a ⇒ a = 256 − 255 = 1.
Hence, the first term of the G.P. is 1.
How many terms of the G.P. 3, 32 , 33 , .... are needed to give the sum 120?
Answer
The given series is a G.P. with a = 3 and r = 3.
Let n terms be required to give the sum of 120.
By formula,
S n = a ( r n − 1 ) r − 1 ∴ 120 = 3 ( 3 n − 1 ) 3 − 1 ⇒ 120 = 3 ( 3 n − 1 ) 2 ⇒ 240 = 3 ( 3 n − 1 ) ⇒ 80 = 3 n − 1 ⇒ 81 = 3 n ⇒ 3 4 = 3 n ⇒ n = 4. S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore 120 = \dfrac{3(3^n - 1)}{3 - 1} \\[1em] \Rightarrow 120 = \dfrac{3(3^n - 1)}{2} \\[1em] \Rightarrow 240 = 3(3^n - 1) \\[1em] \Rightarrow 80 = 3^n - 1 \\[1em] \Rightarrow 81 = 3^n \\[1em] \Rightarrow 3^4 = 3^n \\[1em] \Rightarrow n = 4. S n = r − 1 a ( r n − 1 ) ∴ 120 = 3 − 1 3 ( 3 n − 1 ) ⇒ 120 = 2 3 ( 3 n − 1 ) ⇒ 240 = 3 ( 3 n − 1 ) ⇒ 80 = 3 n − 1 ⇒ 81 = 3 n ⇒ 3 4 = 3 n ⇒ n = 4.
Hence, the required number of terms of the G.P. are 4.
How many terms of the G.P. 1, 4, 16, ... must be taken to have their sum equal to 341?
Answer
The above series is a G.P. with a = 1 and r = 4.
Let n terms be required to give the sum of 341.
By formula,
S n = a ( r n − 1 ) r − 1 ∴ 341 = 1 ( 4 n − 1 ) 4 − 1 ⇒ 341 = ( 4 n − 1 ) 3 ⇒ 341 × 3 = ( 4 n − 1 ) ⇒ 1023 = 4 n − 1 ⇒ 4 n = 1024 ⇒ 4 n = 4 5 ∴ n = 5. S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore 341 = \dfrac{1(4^n - 1)}{4 - 1} \\[1em] \Rightarrow 341 = \dfrac{(4^n - 1)}{3} \\[1em] \Rightarrow 341 \times 3 = (4^n - 1) \\[1em] \Rightarrow 1023 = 4^n - 1 \\[1em] \Rightarrow 4^n = 1024 \\[1em] \Rightarrow 4^n = 4^5 \\[1em] \therefore n = 5. S n = r − 1 a ( r n − 1 ) ∴ 341 = 4 − 1 1 ( 4 n − 1 ) ⇒ 341 = 3 ( 4 n − 1 ) ⇒ 341 × 3 = ( 4 n − 1 ) ⇒ 1023 = 4 n − 1 ⇒ 4 n = 1024 ⇒ 4 n = 4 5 ∴ n = 5.
Hence, the required number of terms of the G.P. are 5.
How many terms of the series,
2 9 − 1 3 + 1 2 + . . . . \dfrac{2}{9} - \dfrac{1}{3} + \dfrac{1}{2} + .... 9 2 − 3 1 + 2 1 + .... 55 72 \dfrac{55}{72} 72 55
Answer
The above series is a G.P. with a = 2 9 and r = − 1 3 2 9 = − 1 3 × 9 2 = − 3 2 \dfrac{2}{9} \text{ and } r = \dfrac{-\dfrac{1}{3}}{\dfrac{2}{9}} = -\dfrac{1}{3} \times \dfrac{9}{2} = -\dfrac{3}{2} 9 2 and r = 9 2 − 3 1 = − 3 1 × 2 9 = − 2 3
Let n terms be required to give the sum of 55 72 \dfrac{55}{72} 72 55
By formula,
S n = a ( r n − 1 ) r − 1 ∴ 55 72 = 2 9 [ ( − 3 2 ) n − 1 ] − 3 2 − 1 ⇒ 55 72 = 2 [ ( − 3 2 ) n − 1 ] 9 × ( − 3 − 2 2 ) ⇒ 55 72 = 4 [ ( − 3 2 ) n − 1 ] 9 × ( − 5 ) ⇒ 55 72 = 4 [ ( − 3 2 ) n − 1 ] − 45 ⇒ 55 × − 45 72 × 4 = ( − 3 2 ) n − 1 ⇒ − 2475 288 = ( − 3 2 ) n − 1 ⇒ − 2475 288 + 1 = ( − 3 2 ) n ⇒ − 2475 + 288 288 = ( − 3 2 ) n ⇒ ( − 3 2 ) n = − 2187 288 S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore \dfrac{55}{72} = \dfrac{\dfrac{2}{9}\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{-\dfrac{3}{2} - 1} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{2\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{9 \times \Big(\dfrac{-3 - 2}{2}\Big)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{4\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{9 \times (-5)} \\[1em] \Rightarrow \dfrac{55}{72} = \dfrac{4\Big[\Big(-\dfrac{3}{2}\Big)^n - 1\Big]}{-45} \\[1em] \Rightarrow \dfrac{55 \times -45}{72 \times 4} = \Big(-\dfrac{3}{2}\Big)^n - 1 \\[1em] \Rightarrow -\dfrac{2475}{288} = \Big(-\dfrac{3}{2}\Big)^n - 1 \\[1em] \Rightarrow -\dfrac{2475}{288} + 1 = \Big(-\dfrac{3}{2}\Big)^n \\[1em] \Rightarrow \dfrac{-2475 + 288}{288} = \Big(-\dfrac{3}{2}\Big)^n \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{2187}{288} \\[1em] S n = r − 1 a ( r n − 1 ) ∴ 72 55 = − 2 3 − 1 9 2 [ ( − 2 3 ) n − 1 ] ⇒ 72 55 = 9 × ( 2 − 3 − 2 ) 2 [ ( − 2 3 ) n − 1 ] ⇒ 72 55 = 9 × ( − 5 ) 4 [ ( − 2 3 ) n − 1 ] ⇒ 72 55 = − 45 4 [ ( − 2 3 ) n − 1 ] ⇒ 72 × 4 55 × − 45 = ( − 2 3 ) n − 1 ⇒ − 288 2475 = ( − 2 3 ) n − 1 ⇒ − 288 2475 + 1 = ( − 2 3 ) n ⇒ 288 − 2475 + 288 = ( − 2 3 ) n ⇒ ( − 2 3 ) n = − 288 2187
Dividing the numerator and denominator by 9
⇒ ( − 3 2 ) n = − 243 32 ⇒ ( − 3 2 ) n = ( − 3 2 ) 5 ∴ n = 5. \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = -\dfrac{243}{32} \\[1em] \Rightarrow \Big(-\dfrac{3}{2}\Big)^n = \Big(-\dfrac{3}{2}\Big)^5 \\[1em] \therefore n = 5. ⇒ ( − 2 3 ) n = − 32 243 ⇒ ( − 2 3 ) n = ( − 2 3 ) 5 ∴ n = 5.
Hence, the required number of terms of the G.P. are 5.
The 2nd and 5th terms of a geometric series are − 1 2 and 1 16 -\dfrac{1}{2} \text{ and } \dfrac{1}{16} − 2 1 and 16 1
Answer
Given,
a 2 = − 1 2 a_2 = -\dfrac{1}{2} a 2 = − 2 1 a 5 = 1 16 a_5 = \dfrac{1}{16} a 5 = 16 1
By formula, a n = a r n − 1 a_n = ar^{n - 1} a n = a r n − 1
⇒ a 2 = a r and a 5 = a r 4 \Rightarrow a_2 = ar \text{ and } a_5 = ar^4 ⇒ a 2 = a r and a 5 = a r 4
Dividing a 5 a_5 a 5 a 2 a_2 a 2
⇒ a r 4 a r = 1 16 − 1 2 ⇒ r 3 = − 1 8 ⇒ r 3 = ( − 1 2 ) 3 ∴ r = − 1 2 . \Rightarrow \dfrac{ar^4}{ar} = \dfrac{\dfrac{1}{16}}{-\dfrac{1}{2}} \\[1em] \Rightarrow r^3 = -\dfrac{1}{8} \\[1em] \Rightarrow r^3 = \Big(-\dfrac{1}{2}\Big)^3 \\[1em] \therefore r = -\dfrac{1}{2}. ⇒ a r a r 4 = − 2 1 16 1 ⇒ r 3 = − 8 1 ⇒ r 3 = ( − 2 1 ) 3 ∴ r = − 2 1 .
Since, a 2 = a r a_2 = ar a 2 = a r
⇒ − 1 2 = a ( − 1 2 ) ⇒ a = 1. \Rightarrow -\dfrac{1}{2} = a\Big(-\dfrac{1}{2}\Big) \\[1em] \Rightarrow a = 1. ⇒ − 2 1 = a ( − 2 1 ) ⇒ a = 1.
By formula,
S n = a ( r n − 1 ) r − 1 ∴ S 8 = 1 [ ( − 1 2 ) 8 − 1 ] − 1 2 − 1 = 1 256 − 1 − 1 − 2 2 = 1 − 256 256 − 3 2 = − 255 × 2 256 × ( − 3 ) = − 510 − 768 S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_8 = \dfrac{1\Big[\Big(-\dfrac{1}{2}\Big)^8 - 1\Big]}{-\dfrac{1}{2} - 1} \\[1em] = \dfrac{\dfrac{1}{256} - 1}{\dfrac{-1 - 2}{2}} \\[1em] = \dfrac{\dfrac{1 - 256}{256}}{\dfrac{-3}{2}} \\[1em] = \dfrac{-255 \times 2}{256 \times (-3)} \\[1em] = \dfrac{-510}{-768} S n = r − 1 a ( r n − 1 ) ∴ S 8 = − 2 1 − 1 1 [ ( − 2 1 ) 8 − 1 ] = 2 − 1 − 2 256 1 − 1 = 2 − 3 256 1 − 256 = 256 × ( − 3 ) − 255 × 2 = − 768 − 510
Dividing numerator and denominator by 6,
= 85 128 . = \dfrac{85}{128}. = 128 85 .
Hence, the sum of the series upto 8 terms is 85 128 \dfrac{85}{128} 128 85
The first term of a G.P. is 27 and 8th term is 1 81 . \dfrac{1}{81}. 81 1 .
Answer
Given, a = 27 and a 8 = 1 81 . a_8 = \dfrac{1}{81}. a 8 = 81 1 .
By formula,
a n = a r n − 1 ∴ a 8 = a r 7 ⇒ 1 81 = 27 r 7 ⇒ 1 81 × 27 = r 7 ⇒ 1 3 4 × 3 3 = r 7 ⇒ r 7 = 1 3 7 ⇒ r = 1 3 . a_n = ar^{n - 1} \\[1em] \therefore a_8 = ar^7 \\[1em] \Rightarrow \dfrac{1}{81} = 27r^7 \\[1em] \Rightarrow \dfrac{1}{81 \times 27} = r^7 \\[1em] \Rightarrow \dfrac{1}{3^4 \times 3^3} = r^7 \\[1em] \Rightarrow r^7 = \dfrac{1}{3^7} \\[1em] \Rightarrow r = \dfrac{1}{3}. a n = a r n − 1 ∴ a 8 = a r 7 ⇒ 81 1 = 27 r 7 ⇒ 81 × 27 1 = r 7 ⇒ 3 4 × 3 3 1 = r 7 ⇒ r 7 = 3 7 1 ⇒ r = 3 1 .
By formula,
S n = a ( r n − 1 ) r − 1 ∴ S 10 = 27 [ ( 1 3 ) 10 − 1 ] 1 3 − 1 = 27 ( 1 3 10 − 1 ) 1 − 3 3 = 27 ( 1 − 3 10 ) 3 10 − 2 3 = 3 3 ( 1 − 3 10 ) 3 10 − 2 × 3 = 3 4 ( 1 − 3 10 ) − 2 × 3 10 = 3 4 2 × − ( 1 − 3 10 ) 3 10 = 81 2 ( 3 10 − 1 3 10 ) = 81 2 ( 1 − 1 3 10 ) . S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S_{10} = \dfrac{27\Big[\Big(\dfrac{1}{3}\Big)^{10} - 1\Big]}{\dfrac{1}{3} - 1} \\[1em] = \dfrac{27(\dfrac{1}{3^{10}} - 1)}{\dfrac{1 - 3}{3}} \\[1em] = \dfrac{27 \dfrac{(1 - 3^{10})}{3^{10}}}{-\dfrac{2}{3}} \\[1em] = \dfrac{3^3 \dfrac{(1 - 3^{10})}{3^{10}}}{-2} \times 3 \\[1em] = \dfrac{3^4(1 - 3^{10})}{-2 \times 3^{10}} \\[1em] = \dfrac{3^4}{2} \times \dfrac{-(1 - 3^{10})}{3^{10}} \\[1em] = \dfrac{81}{2} \Big(\dfrac{3^{10} - 1}{3^{10}}\Big) \\[1em] = \dfrac{81}{2} \Big(1 - \dfrac{1}{3^{10}}\Big). S n = r − 1 a ( r n − 1 ) ∴ S 10 = 3 1 − 1 27 [ ( 3 1 ) 10 − 1 ] = 3 1 − 3 27 ( 3 10 1 − 1 ) = − 3 2 27 3 10 ( 1 − 3 10 ) = − 2 3 3 3 10 ( 1 − 3 10 ) × 3 = − 2 × 3 10 3 4 ( 1 − 3 10 ) = 2 3 4 × 3 10 − ( 1 − 3 10 ) = 2 81 ( 3 10 3 10 − 1 ) = 2 81 ( 1 − 3 10 1 ) .
Hence, the sum of first 10 terms of G.P. is 81 2 ( 1 − 1 3 10 ) . \dfrac{81}{2} \Big(1 - \dfrac{1}{3^{10}}\Big). 2 81 ( 1 − 3 10 1 ) .
Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728.
Answer
Given, r = 3, l = 486 and Sn = 728.
Let the last term be nth term.
Hence, l = an = 486.
Using formula,
a n = a r n − 1 ⇒ 486 = a ( 3 ) n − 1 ⇒ a 3 n 3 = 486 ⇒ 3 n a = 486 × 3 ⇒ 3 n a = 1458 ⇒ a = 1458 3 n . (Eq 1) a_n = ar^{n - 1} \\[1em] \Rightarrow 486 = a(3)^{n - 1} \\[1em] \Rightarrow a\dfrac{3^n}{3} = 486 \\[1em] \Rightarrow 3^na = 486 \times 3 \\[1em] \Rightarrow 3^na = 1458 \\[1em] \Rightarrow a = \dfrac{1458}{3^n}. \text{ (Eq 1)} a n = a r n − 1 ⇒ 486 = a ( 3 ) n − 1 ⇒ a 3 3 n = 486 ⇒ 3 n a = 486 × 3 ⇒ 3 n a = 1458 ⇒ a = 3 n 1458 . (Eq 1)
Given, S n = 728 S_n = 728 S n = 728
Putting value of a = 1458 3 n a = \dfrac{1458}{3^n} a = 3 n 1458 S n = a ( r n − 1 ) r − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} S n = r − 1 a ( r n − 1 )
⇒ S n = 1458 3 n ( 3 n − 1 ) 3 − 1 ⇒ 728 = 1458 − 1458 3 n 2 ⇒ 1458 − 1458 3 n = 728 × 2 ⇒ 1458 3 n = 1458 − 1456 ⇒ 1458 3 n = 2 ⇒ 3 n = 1458 2 ⇒ 3 n = 729 \Rightarrow S_n = \dfrac{\dfrac{1458}{3^n}(3^n - 1)}{3 - 1} \\[1em] \Rightarrow 728 = \dfrac{1458 - \dfrac{1458}{3^n}}{2} \\[1em] \Rightarrow 1458 - \dfrac{1458}{3^n} = 728 \times 2 \\[1em] \Rightarrow \dfrac{1458}{3^n} = 1458 - 1456 \\[1em] \Rightarrow \dfrac{1458}{3^n} = 2 \\[1em] \Rightarrow 3^n = \dfrac{1458}{2} \\[1em] \Rightarrow 3^n = 729 ⇒ S n = 3 − 1 3 n 1458 ( 3 n − 1 ) ⇒ 728 = 2 1458 − 3 n 1458 ⇒ 1458 − 3 n 1458 = 728 × 2 ⇒ 3 n 1458 = 1458 − 1456 ⇒ 3 n 1458 = 2 ⇒ 3 n = 2 1458 ⇒ 3 n = 729
Putting the above value in Eq 1,
⇒ a = 1458 729 ∴ a = 2. \Rightarrow a = \dfrac{1458}{729} \\[1em] \therefore a = 2. ⇒ a = 729 1458 ∴ a = 2.
Hence, the first term of the G.P. is 2.
In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.
Answer
Let nth term be the last term of the G.P.
Given, a = 7, l = an = 448 and Sn = 889.
Using formula,
a n = a r n − 1 ⇒ 448 = 7 ( r ) n − 1 ⇒ r n − 1 = 64 ⇒ r n r = 64 ⇒ r n = 64 r . (Eq 1) a_n = ar^{n - 1} \\[1em] \Rightarrow 448 = 7(r)^{n - 1} \\[1em] \Rightarrow r^{n - 1} = 64 \\[1em] \Rightarrow \dfrac{r^n}{r} = 64 \\[1em] \Rightarrow r^n = 64r. \text{ (Eq 1)} \\[1em] a n = a r n − 1 ⇒ 448 = 7 ( r ) n − 1 ⇒ r n − 1 = 64 ⇒ r r n = 64 ⇒ r n = 64 r . (Eq 1)
Given, Sn = 889
Putting value of rn from Eq 1 in the formula S n = a ( r n − 1 ) r − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} S n = r − 1 a ( r n − 1 )
⇒ 889 = 7 ( 64 r − 1 ) r − 1 ⇒ 889 ( r − 1 ) = 448 r − 7 ⇒ 889 r − 889 = 448 r − 7 ⇒ 889 r − 448 r = 889 − 7 ⇒ 441 r = 882 ∴ r = 2. \Rightarrow 889 = \dfrac{7(64r - 1)}{r - 1} \\[1em] \Rightarrow 889(r - 1) = 448r - 7 \\[1em] \Rightarrow 889r - 889 = 448r - 7 \\[1em] \Rightarrow 889r - 448r = 889 - 7 \\[1em] \Rightarrow 441r = 882 \\[1em] \therefore r = 2. ⇒ 889 = r − 1 7 ( 64 r − 1 ) ⇒ 889 ( r − 1 ) = 448 r − 7 ⇒ 889 r − 889 = 448 r − 7 ⇒ 889 r − 448 r = 889 − 7 ⇒ 441 r = 882 ∴ r = 2.
Hence, the common ratio of the G.P. is 2.
Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.
Answer
Given, r = 3 and S7 = 2186.
Using formula S n = a ( r n − 1 ) r − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} S n = r − 1 a ( r n − 1 )
⇒ S 7 = a ( 3 7 − 1 ) 3 − 1 ⇒ 2186 = a ( 2187 − 1 ) 2 ⇒ 2186 a 2 = 2186 ⇒ a = 2. \Rightarrow S_7 = \dfrac{a(3^7 - 1)}{3 - 1} \\[1em] \Rightarrow 2186 = \dfrac{a(2187 - 1)}{2} \\[1em] \Rightarrow \dfrac{2186a}{2} = 2186 \\[1em] \Rightarrow a = 2. \\[1em] ⇒ S 7 = 3 − 1 a ( 3 7 − 1 ) ⇒ 2186 = 2 a ( 2187 − 1 ) ⇒ 2 2186 a = 2186 ⇒ a = 2.
∴ Third term = a3 = ar2 = 2(3)2 = 18.
Hence, the third term of the G.P. is 18.
If the first term of a G.P. is 5 and the sum of first three terms is 31 5 , \dfrac{31}{5}, 5 31 ,
Answer
Given, a = 5 and S3 = 31 5 \dfrac{31}{5} 5 31
Using formula S n = a ( r n − 1 ) r − 1 S_n = \dfrac{a(r^n - 1)}{r - 1} S n = r − 1 a ( r n − 1 )
⇒ S 3 = 5 ( r 3 − 1 ) r − 1 ⇒ 31 5 = 5 ( r 3 − 1 ) r − 1 ⇒ 31 ( r − 1 ) = 25 ( r 3 − 1 ) ⇒ 25 ( r − 1 ) ( r 2 + r + 1 ) = 31 ( r − 1 ) \Rightarrow S_3 = \dfrac{5(r^3 - 1)}{r - 1} \\[1em] \Rightarrow \dfrac{31}{5} = \dfrac{5(r^3 - 1)}{r - 1} \\[1em] \Rightarrow 31(r - 1) = 25(r^3 - 1) \\[1em] \Rightarrow 25(r - 1)(r^2 + r + 1) = 31(r - 1) \\[1em] ⇒ S 3 = r − 1 5 ( r 3 − 1 ) ⇒ 5 31 = r − 1 5 ( r 3 − 1 ) ⇒ 31 ( r − 1 ) = 25 ( r 3 − 1 ) ⇒ 25 ( r − 1 ) ( r 2 + r + 1 ) = 31 ( r − 1 )
Dividing by (r - 1) on both sides we get,
⇒ 25 ( r 2 + r + 1 ) = 31 ⇒ 25 r 2 + 25 r + 25 = 31 ⇒ 25 r 2 + 25 r + 25 − 31 = 0 ⇒ 25 r 2 + 25 r − 6 = 0 ⇒ 25 r 2 + 30 r − 5 r − 6 = 0 ⇒ 5 r ( 5 r + 6 ) − 1 ( 5 r + 6 ) = 0 ⇒ ( 5 r − 1 ) ( 5 r + 6 ) = 0 ⇒ 5 r − 1 = 0 or 5 r + 6 = 0 ⇒ 5 r = 1 or 5 r = − 6 ⇒ r = 1 5 or r = − 6 5 . \Rightarrow 25(r^2 + r + 1) = 31 \\[1em] \Rightarrow 25r^2 + 25r + 25 = 31 \\[1em] \Rightarrow 25r^2 + 25r + 25 - 31 = 0 \\[1em] \Rightarrow 25r^2 + 25r - 6 = 0 \\[1em] \Rightarrow 25r^2 + 30r - 5r - 6 = 0 \\[1em] \Rightarrow 5r(5r + 6) - 1(5r + 6) = 0 \\[1em] \Rightarrow (5r - 1)(5r + 6) = 0 \\[1em] \Rightarrow 5r - 1 = 0 \text{ or } 5r + 6 = 0 \\[1em] \Rightarrow 5r = 1 \text{ or } 5r = -6 \\[1em] \Rightarrow r = \dfrac{1}{5} \text{ or } r = -\dfrac{6}{5}. ⇒ 25 ( r 2 + r + 1 ) = 31 ⇒ 25 r 2 + 25 r + 25 = 31 ⇒ 25 r 2 + 25 r + 25 − 31 = 0 ⇒ 25 r 2 + 25 r − 6 = 0 ⇒ 25 r 2 + 30 r − 5 r − 6 = 0 ⇒ 5 r ( 5 r + 6 ) − 1 ( 5 r + 6 ) = 0 ⇒ ( 5 r − 1 ) ( 5 r + 6 ) = 0 ⇒ 5 r − 1 = 0 or 5 r + 6 = 0 ⇒ 5 r = 1 or 5 r = − 6 ⇒ r = 5 1 or r = − 5 6 .
Hence, the common ratio of the G.P. is 1 5 or − 6 5 . \dfrac{1}{5} \text{ or } -\dfrac{6}{5}. 5 1 or − 5 6 .
In a Geometric Progression (G.P.) the first term is 24 and the fifth term is 8. Find the ninth term of the G.P.
Answer
Let first term of G.P. be a and common ratio be r.
Given,
First term (a) = 24
Fifth term (ar4 ) = 8
⇒ a r 4 a = 8 24 ⇒ r 4 = 1 3 ⇒ ( r 4 ) 2 = ( 1 3 ) 2 ⇒ r 8 = 1 9 . \Rightarrow \dfrac{ar^4}{a} = \dfrac{8}{24} \\[1em] \Rightarrow r^4 = \dfrac{1}{3} \Rightarrow (r^4)^2 = \Big(\dfrac{1}{3}\Big)^2 \\[1em] \Rightarrow r^8 = \dfrac{1}{9}. ⇒ a a r 4 = 24 8 ⇒ r 4 = 3 1 ⇒ ( r 4 ) 2 = ( 3 1 ) 2 ⇒ r 8 = 9 1 .
By formula,
Ninth term of G.P. (a9 ) = ar8
= 24 × 1 9 = 8 3 24 \times \dfrac{1}{9} = \dfrac{8}{3} 24 × 9 1 = 3 8
Hence, ninth term of G.P. = 8 3 \dfrac{8}{3} 3 8
