Question

In a geometric progression, the sum of the first $3$ terms is $7$ and the sum of the next $3$ terms is $56$. Find the geometric progression

Answer 1

Let $a_n$ denote $n^{th}$ term of geometric progression.

Given : $a_1+a_2+a_3=7$ and $a_4+a_5+a_6=56$

Since $a_n=a{r}^{n-1}$

$⟹$ $a+ar+a{r}^2=7$ ....... $\left(1\right)$ and $a{r}^3+a{r}^4+a{r}^5=56$ ......... $\left(2\right)$

In $\left(2\right)$

$r^3(a+ar+a{r}^2)=56$

$⟹$ $r^3\left(7\right)=56$ .......... from $\left(1\right)$

$⟹$ $r^3=8$

$\therefore$ $r=2$

Substituting the value of $r$ in $\left(1\right)$ we get $a=1$

$\therefore$ $r=2$ and $a=1$

Hence, the terms in geometric progression are $1,2,4,8,16,32,...........$