Question

The difference between the fourth and the first term of a geometric progression is $52$ and the sum of the first three terms is $26$. Calculate the sum of the first six terms of the progression.

Answer 1

Let the terms be $T_1,T_2,T_3\&T_4$

$\therefore T_1=a_1,T_2=a_{1}r,T_3=a_1{r}^2,T_4=a_1{r}^3$

Now,$T_4-T_1=52$

$\Rightarrow a_1{r}^3-a_1=52\Rightarrow a_1(r^3-1)=52\Rightarrow a_1(r-1)(r^2+r+1)=52-\left(i\right)T_1+T_2+T_3=26\Rightarrow a_1+a_{1}r+a_1{r}^2=26\Rightarrow a_1(1+r+r^2)=26-\left(ii\right)\therefore \left(i\right)÷\left(ii\right)\Rightarrow \frac{a_1(r-1)(r^2+r+1)}{a_1(1+r+r^2)}=\frac{52}{26}\Rightarrow r-1=2\Rightarrow r=3From\left(i\right)a_1(3-1)(9+3+1)=52\Rightarrow a_1\times 2\times 3=52$

$\therefore$ The first six terms are.

$\Rightarrow a_1=\frac{52}{2\times 13}=2=2,2\times 3,2\times 3^2,2\times 3^3,2\times 3^4,2\times 3^5=2,6,18,54,162,486$