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

Solution:

(d) $3 / 4$

Let the terms of the G.P. be $a, a_{2}, a_{3}, a_{4}, a_{5}, \ldots, \infty$.

And, let the common ratio be $r$.

Now, $a+a_{2}=1$

$\therefore a+a r=1 \quad \ldots \ldots$ (i)

Also, $a=2\left(a_{2}+a_{3}+a_{4}+a_{5}+\ldots \infty\right)$

$\Rightarrow a=2\left(a r+a r^{2}+a r^{3}+a r^{4}+\ldots \infty\right)$

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

$\Rightarrow 1-r=2 r$

$\Rightarrow 3 r=1$

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

Putting the value of $\mathrm{r}$ in $(\mathrm{i})$ :

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

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

$\Rightarrow 4 a=3$

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