The 2 n d and 5 t h terms of a geometric series are - 1 2 and 1 16 respectively. Find the sum of the series upto 8 terms.

let a be the first term and r be the common ratio of the given G . P . Given that, a 2 = - 1 2 We know that, n t h term of a G . P . is a n = a r n - 1 . a r = - 1 2 (i) And a 5 = 1 16 ⇒ a r 4 = 1 16 (ii) On dividing equation (ii) by equation (i), we get r 3 = 1 16 - 1 2 = - 1 8 = - 1 2 3 ⇒ r = - 1 2 ⇒ | r | = 1 2 < 1 From (i), a × - 1 2 = - 1 2 ⇒ a = 1 We know that, sum of n terms in a G . P . is S n = a 1 - r n 1 - r . ∴ S 8 = 1 1 - - 1 2 8 1 - - 1 2 = 2 3 1 - 1 256 = 2 × 255 3 × 256 = 85 128 ∴ The sum of the given series upto 8 terms is 85 128 .

The first term of a G . P . is 27 and 8 t h term is 1 81 . Find the sum of its first 10 terms.

Let a be the first term and r be the common ratio of the given G . P . Here, a = 27 and a 8 = 1 81 We know that, n t h term of a G . P . is a n = a r n - 1 . Now, a r 7 = 1 81 ⇒ 27 r 7 = 1 81 ⇒ r 7 = 1 27 × 81 = 1 3 3 × 3 4 = 1 3 7 ⇒ r = 1 3 < 1 We know that, sum of n terms in a G . P . is S n = a 1 - r n 1 - r ∴ S 10 = 27 1 - 1 3 10 1 - 1 3 = 81 2 1 - 1 3 10 ∴ The sum of its first 10 terms of the given series is 81 2 1 - 1 3 10 .

If the first term of a G . P . is 5 and the sum of first three terms is 31 5 , find the common ratio.

Let a be the first term and r be the common ratio of the given G . P . respectively. Given that, a = 5 and S 3 = 31 5 . We know that, sum of n terms of a G . P . is S n = a 1 - r n 1 - r . Now, S 3 = 5 l - r 3 1 - r ⇒ 31 5 = 5 1 - r 3 1 - r ⇒ 21 25 = 1 + r 2 + r [ ∵ r ≠ 1 ] ⇒ 25 r 2 + 25 r - 6 = 0 ⇒ 25 r 2 + 30 r - 5 r - 6 = 0 ⇒ ( 5 r + 6 ) ( 5 r - 1 ) = 0 ⇒ r = - 6 5 , 1 5 Hence, the required common ratio of G . P . is - 6 5 or 1 5 .

Evaluate ∑ n = 1 50 2 n - 1 .

Given that, the expression is ∑ n = 1 50 2 n - 1 = ∑ n = 1 50 2 n - ∑ n = 1 50 1 = 2 1 + 2 2 + 2 3 + … . upto 50 terms - 1 + 1 + 1 … upto 50 terms which forms a G . P . with first term a = 2 We know that, common ratio of the G . P . is r = T n T n - 1 . Common ratio of the G . P . r = 2 > 1 We know that, sum of n terms of the given G . P . is S n = a r n - 1 r - 1 , r > 1 . Now, 1 = 2 2 50 - 1 2 - 1 - 50 = 2 51 - 52 ∴ The value of the given expression is 2 51 - 52 .