Question

The difference between the third and the second term of a geometric progression is $12$. If we add $10$ to the first term and 8 to the second and leave the third term unchanged, the now three numbers will form an arithmetic progression. Find the sum of the first five terms of the geometric progression.

Answer 1

Let the terms be, $a,ar,a{r}^2,a{r}^3,..........$

Given,

$a{r}^2-ar=12$

$ar(r-1)=12$.........(1)

$a+10,ar+8,a{r}^2$ are in A.P.

$\Rightarrow ar+8-a-10=a{r}^2-ar-8$

From (1)

$ar-a-2=12-8=4$

$a(r-1)=6\Rightarrow r-1=\frac{6}{a}$

From (1)

$ar\left(\frac{6}{a}\right)=12$

$\Rightarrow r=2$

$a=\frac{6}{2-1}$

$\Rightarrow a=6$

Sum of 5 terms,

$=a+ar+a{r}^2+a{r}^3+a{r}^4$

$=6+6\left(2\right)+6\left(2^2\right)+6\left(2^3\right)+6\left(2^4\right)$

$=186$