Question

The fourth term of a geometric progression exceeds the second term by $24$ and the sum of the second and the third term is $6$. Find the progression.

Answer 1

In a geometric progression,

$T_n=a_1{n}^{n-1}$

where $a_1$is the first term & $n$ is the common ratio.

$T_4-T_2=24\Rightarrow a_1{n}^3-a_{1}n=24\Rightarrow a_{1}r(r^2-1)=24\Rightarrow a_{1}r(r+1)(r-1)=24-\left(i\right)$

Also, $T_2+T_3=6$

$\Rightarrow a_{1}r+a_1{r}^2=6\Rightarrow a_{1}r(1+r)=6-\left(ii\right)$

$\therefore$ Dividing $\left(1\right)$ by $\left(ii\right)$

$\frac{a_{1}r(r+1)(r-1)}{a_{1}r(1+r)}=\frac{24}{6}\Rightarrow r-1=4\Rightarrow r=5$

From $\left(ii\right)$

$a_1\times 5(1+5)=6\Rightarrow a_1\times 5\times 6=6\Rightarrow a_1=\frac{6}{5\times 6}=\frac{1}{5}$

$\therefore$ The progression is $\frac{1}{5},1,5,25,125,.....$