# CLASS-10GEOMETRICAL PROGRESSION - SUM OF e-TERMS OF GP

Sum of n-terms of a G.P.

Theorem 3.)  Prove that the sum of ‘e’ terms of a G.P. with first term ‘a’ and common ratio ‘r’ is given by –

Sₑ = ea, when  r = 1,

a (1 - rᵉ)

Sₑ = ------------- , when  r < 1

(1 – r)

a (rᵉ - 1)

Sₑ = --------------, when  r > 1

(r – 1)

Proof - Let us consider a G.P. with first term ‘a’ and common ratio ‘r’. Then,

Sₑ = a + ar + ar² + ………………..+ arᵉˉ¹ ……………………(i)

Case 1.   If r = 1, we have

Sₑ = a + a + a + ……………… to e terms = ea

Case 2.   If r ≠ 1, we have

rSₑ = ar + ar² + ar³ + ……………… + arᵉˉ¹ + arᵉ

On subtracting (ii) from (i), we get –

(1 – r)Sₑ = (a - arᵉ) = a (1 - rᵉ) ………………(ii)

a (1 - rᵉ)                       a (rᵉ - 1)

=> Sₑ = ------------     or    Sₑ = ------------

(1 – r)                          (r – 1)

a (1 - rᵉ)

Sₑ = ------------, when   r < 1

(1 – r)

a (rᵉ - 1)

Sₑ = ------------, when   r > 1

(r – 1)

Remarks :- If a G.P. contains ‘e’ terms with first term = a, common ratio = r, and last term = l, then –

l = arᵉˉ¹ …………………..(i)

Case.1) -  When r < 1, we have

a (1 - rᵉ)          (a - arᵉ)

Sₑ = ------------ = -----------

(1 – r)           (1 – r)

Example.1) Find the sum of 9 terms of the G.P. 3, 6, 12, 24,………………

Ans.) The given G.P. is 3, 6, 12, 24,………………

6

Here, a = 3, r = ------- = 2 > 1 and e = 9

3

Using the following formula, and we get -

a (rᵉ - 1)

Sₑ = ------------, when  r > 1

(r – 1)

3 X (2⁹ - 1)

S₉ = ---------------

(2 – 1)

3 X (512 – 1)

= ----------------

1

=  (3 X 511) = 1533            (Ans.)