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.)
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