Find the sum of the series 81 - 27 + 9 - … - 1 27 .

Given that, the series is 81 - 27 + 9 - … - 1 27 . It forms a G . P . with first term a = 81 and I = - 1 27 . We know that, common ratio of the G . P . is r = T n T n - 1 . Common ratio r = - 27 81 = - 1 3 We know that, Sum of n terms in a G . P . is S n = a 1 - r n 1 - r . Now, S n = a - a r n 1 - r = a - a r n - 1 · r 1 - r = a - l r 1 - r ∵ l = a n = a r n - 1 ∴ S n = 81 - - 1 27 - 1 3 1 - - 1 3 = 81 - 1 81 4 3 = ( 81 ) 2 - 1 2 × 3 81 × 4 = 80 × 82 × 3 81 × 4 = 1640 27 ∴ The sum of first ten terms of the given series is 1640 27 .

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 .