Question

The sum of the first and the third term of a geometric progression is $20$ and the sum of its first three terms is $26$. Find the progression.

Answer 1

Let the first term be $a$ & common ratio be $r$.

$\therefore T_1=a{T}_2=ar{T}_3=a{r}^2\therefore a+a{r}^2=20\Rightarrow a(1+r^2)=20-\left(i\right)S_3=26\Rightarrow \frac{a(1-r^2)}{1-r}=26\Rightarrow \frac{a(1+r)(1-r)}{1-r}=26\Rightarrow a(1+r)=26-\left(ii\right)$

Also,we know that $T_1+T_3=20$ & $T_1+T_2+T_3=26$

$\therefore (T_1+T_2+T_3)-(T_1+T_3)=26-20\Rightarrow T_2=6\Rightarrow ar=6[T_2=ar]-\left(iii\right)$

Now from $\left(ii\right)$

$a(1+r)=26\Rightarrow a+ar=26$

Putting $ar=6$ from $\left(iii\right)$

$\Rightarrow a=20\therefore 1^{st}term=202^{nd}term=6\Rightarrow ar=6\Rightarrow 20\times r=6r=\frac{6}{20}=\frac{3}{10}$

Hence,the progression is $20,6,\frac{18}{10},\frac{54}{100}$