Question

The difference between the first and the fifth term of a geometric progression, whose all terms are positive numbers is $15$ and the sum of the first and the third term of the progression is $20$. Calculate the sum of the first five terms of the progression.

Answer 1

Let the series be $a,ar,a{r}^2......$ where $a$ is the first term and $r$ is the common ratio.

$\therefore T_1-T_5=15\Rightarrow a-a{r}^4=15\Rightarrow a(1-r^4)=15-\left(i\right)$

Also, $T_1+T_3=20$

$\Rightarrow a+a{r}^2=20\Rightarrow a(1+r^2)=20-\left(ii\right)$

Hence, dividing $\left(i\right)$ by $\left(ii\right)$

$\frac{a(1-r^4)}{a(1+r^2)}=\frac{15}{20}\Rightarrow \frac{(1-r^2)(1+r^2)}{(1+r^2)}=\frac{15}{20}\Rightarrow 1-r^2=\frac{3}{4}\Rightarrow r^2=1-\frac{3}{4}=\frac{1}{4}\Rightarrow r=\pm \frac{1}{4}$

Since,all terms are positive,$r=\frac{1}{4}$

From $\left(i\right)$

$a(1-\frac{1}{4^4})=15\Rightarrow a(1-\frac{1}{256})=15\Rightarrow a=\frac{15\times 256}{255}=\frac{256}{17}{T}_1=\frac{256}{17}{T}_2=\frac{256}{17}\times \frac{1}{4}=\frac{64}{17}{T}_3=\frac{256}{17}\times \frac{1}{16}=\frac{16}{17}{T}_4=\frac{256}{17}\times \frac{1}{64}=\frac{4}{17}{T}_5=\frac{256}{17}\times \frac{1}{256}=\frac{1}{17}$

The terms are $\frac{256}{17},\frac{64}{17},\frac{16}{17},\frac{4}{17},\frac{1}{17}$