Question

If the sum of first two terms of an infinite GP is 1 and every term is twice the sum of all the successive terms, then its first term is

Answer 1

(d) $\frac{3}{4}$

Let the terms of the G.P. be $a,a_2,a_3,a_4,a_5.......,\infty$

And, let the common ratio be r

Now, $a+a_2=1$

$\therefore a+ar=1$ ........ (i)

Also, $a=2(a_2+a_3+a_4+a_5+.......\infty )$

$\Rightarrow a=2(ar+a{r}^2+a{r}^3+a{r}^4+......\infty )$

$\Rightarrow a=2\left(\frac{ar}{1-r}\right)$

$\Rightarrow 1-r=2r$

$\Rightarrow 3r=1$

$\Rightarrow r=\frac{1}{3}$.

Putting the value of r in (i) :

$a+\frac{a}{3}=1$

$\Rightarrow \frac{4a}{3}=1$

$\Rightarrow 4a=3$

$\Rightarrow a=\frac{3}{4}$