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

' T n ' represents the n th term of a G.P. series. T n = a r n - 1 ⇒ 1 81 = 27 × r 8 - 1 ⇒ 1 81 = 27 × r 7 ⇒ 1 2187 = r 7 ⇒ 1 3 7 = r 7 ∴ r = 1 3 We know that, the sum of a G.P. series is represented by the formula, S n = a r n - 1 r - 1 , when r < 1 . ' S n ' represents the sum of the G.P. series upto n t h terms, ' a ' represents the first term, ' r ' represents the common ratio and ' n ' represents the number of terms. Here, a = 1 r = (ratio between the n term and n - 1 term) 3 ÷ 1 = 3 n = 7 terms ∴ S n = 27 × 1 - 1 3 10 1 - 1 3 ⇒ S n = 27 × 1 - 1 59049 2 3 ⇒ S n = 27 × 59048 59049 2 3 ⇒ S n = 27 × 39524 19683 ∴ S n = 39524 729

The 2 n d and 5 t h terms of a GP are - 1 2 and 1 16 respectively. Find the sum of the GP up to 8 terms.

Given 2 nd term = ar 2 - 1 = a r 1 = - 1 2 5 th term = a r 5 - 1 = a r 4 = 1 16 Dividing the 5 th term using the 3 rd term a r 4 a r = 1 16 - 1 2 r 3 = - 1 8 ∴ r = - 1 2 ⇒ a = 1 We know that, the sum of a G.P. series is represented by the formula, S n = a r n - 1 r - 1 , when r < 1 . ' S n ' represents the sum of the G.P. series upto n t h terms, ' a ' represents the first term, ' r ' represents the common ratio and ' n ' represents the number of terms. n = 8 terms S n = 1 × 1 - - 1 8 2 1 - - 1 2 S n = 1 - 1 250 3 2 ⇒ S n = 255 250 3 2 ∴ S n = 170 256

The 4 t h and 7 t h terms of a GP are 1 27 and 1 729 respectively. Find the sum of n terms of the GP.

4 th term = a r - 1 = a r 3 = 1 27 7 th term = ar 7 - 1 = ar 6 = 1 729 Dividing the 7 th term by the 4 th term, a r 6 a r 3 = 1 729 1 27 ⇒ r 3 = 1 27 … … ( i ) ∴ r = 1 3 a r 3 = 1 27 [Putting from eqn ( i ) ] a 1 27 = 1 27 ∴ a = 1 Sum of a G.P. series is represented by the formula, S n = a 1 - r n 1 - r , when r < 1 . ' S n ' represents the sum of the G.P. series upto n t h terms, ' a ' represents the first term, ' r ' represents the common ratio and ' n ' represents the number of terms. Here, a = 1 r = 1 3 n terms S n = 1 × 1 - 1 3 n 1 - 1 3 ∴ S n = 1 - 1 3 n 2 3 ⇒ S n = 3 2 1 - 1 3 n