The sum of first two terms of an infinite G.P. is 5 and each term is three times the sum of the succeeding terms.

Solution:

Let the first term be a and the common difference be r.

$\therefore a_{1}+a_{2}=5$

$\Rightarrow a+a r=5$          ....(i)

Also, $a_{n}=3\left[a_{n+1}+a_{n+2}+a_{n+3}+\ldots \infty\right] \forall n \in N$

$\Rightarrow a r^{n-1}=3\left[a r^{n+1}+a r^{n+2}+a r^{n+3}+\ldots \infty\right]$

$\Rightarrow a r^{n-1}=\frac{3 a r^{n}}{1-r}$

$\Rightarrow 1-r=3 r$

$\Rightarrow 4 r=1$

$\Rightarrow r=\frac{1}{4}$

Putting $r=\frac{1}{4}$ in (i) :

$a+\frac{a}{4}=5$

$\Rightarrow 5 a=20$

$\Rightarrow a=4$

Thus, the G.P. is $4,1, \frac{1}{4}, \frac{1}{16}, \ldots \infty$.