If S is the sum to infinity of a G.P. whose first term is 1, then the

1.	If S is the sum to infinity of a G.P. whose first term is 1, then the sum of its first n terms is(a) S\(\big(1-\frac{1}{S}\big)^{n-1}\)(b) S \(\big(1-\frac{1}{S}\big)^{n}\)(c) S\(\big\{1-\big(1-\frac{1}{S}\big)^{n-1}\big\}\)(d) S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\)
Answer» (d) S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\) Let the first term of the G.P. be a and common ratio r. Given S = \(\frac{a}{1-r}\) ⇒ S = \(\frac{1}{1-r}\) ⇒ 1 - r = \(\frac{1}{S}\) ⇒ r = 1 - \(\frac{1}{S}\) Let S_n be the sum of first n terms of the series. Then, S_n = \(\frac{1(1-r^n)}{1-r}\) (∵ \| r \| < 1) = \(\frac{1\big(1-\big(1-\frac{1}{S}\big)^n\big)}{\big(1-\big(1-\frac{1}{S}\big)\big)}\) = S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\)

If S is the sum to infinity of a G.P. whose first term is 1, then the sum of its first n terms is(a) S\(\big(1-\frac{1}{S}\big)^{n-1}\)(b) S \(\big(1-\frac{1}{S}\big)^{n}\)(c) S\(\big\{1-\big(1-\frac{1}{S}\big)^{n-1}\big\}\)(d) S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\)

Answer»

(d) S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\)

Let the first term of the G.P. be a and common ratio r.

Given S = \(\frac{a}{1-r}\) ⇒ S = \(\frac{1}{1-r}\)

⇒ 1 - r = \(\frac{1}{S}\) ⇒ r = 1 - \(\frac{1}{S}\)

Let S_n be the sum of first n terms of the series. Then,

S_n = \(\frac{1(1-r^n)}{1-r}\) (∵ | r | < 1)

= \(\frac{1\big(1-\big(1-\frac{1}{S}\big)^n\big)}{\big(1-\big(1-\frac{1}{S}\big)\big)}\) = S\(\big\{1-\big(1-\frac{1}{S}\big)^{n}\big\}\)