Question

The sum of first two terms of an infinite G.P. is 1 and every term is twice the sum of the successive terms. Its first term is :

a) 1/3
b) 2/3
c) 3/4
d) 1/4

________________________

Explanation required.
Thank you..​

Answer 1

AnswEr :

Let a and r be the first term and common ratio here respectively.

We're given an infinite geometric progression, the sequence would be something like this :

$\sf \: a + ar + {ar}^{2} + \dots \dots \dots \: + \infty$

According to the given condition,

$\sf \: a + ar = 1 \longrightarrow(1)$

Every term in the GP is twice the sum of it's successive terms,

$\sf \: a = 2( ar + {ar}^{2} + \dots + {ar}^{n} ) \longrightarrow \: (2)$

Now, the above sequence forms a GP with a common ratio r and first term ar, such that r < 1.

Sum of terms in a GP :

$\boxed{ \boxed{ \sf \: S = \dfrac{a}{1 - {r}^{n} } }}$

Since, sum of successive terms is equal to the preceding term.

$\longrightarrow \sf a = \dfrac{2ar}{1 - r} \\ \\ \longrightarrow \sf \: 1 = \dfrac{2r}{1 - r} \\ \\ \longrightarrow \sf \: 1 - r = 2r \\ \\ \longrightarrow \: \underline{\boxed{ \sf r = \dfrac{1}{3} }}$

Using condition (1),

$\implies \sf \: a + \dfrac{a}{3} = 1 \\ \\ \implies \: \sf \: a = \dfrac{1}{ \big(1 + \dfrac{1}{3} \big)} \\ \\ \implies \boxed{ \boxed{ \sf \: a = \dfrac{3}{4} }}$

Option (C) is the answer.

Answer 2

If the first tern is a and common ratio is r. The infinite sum of the series is $\frac{a}{1 - r}$

The general term is ${ar}^{n}$.The sum of all the successive terms is

$({ar}^{n}) r + ( {ar}^{n} ) {r}^{2} + ({ar}^{n}) {r}^{3}$+.....∞

= $\frac{ ({ar}^{n})r }{(1 -r )}$

∴ ${ar}^{n} = 2\frac{ ({ar}^{n})r }{(1 -r )}$

$1 - r = 2r \\ r = \frac{1}{3}$

Also given that,

$a +a r = a(1 + r) = 1$

So,

$a = \frac{1}{(1 + 6)} = \frac{3}{4}$

$\underline\mathfrak{Answer: \: \: a \: = \frac{3}{4}}$