The difference between the fourth term and the first term of a Geometrical Progresssion is 52. If the sum of…
The difference between the fourth term and the first term of a Geometrical Progresssion is 52. If the sum of its first three terms is 26 , then the sum of the first six terms of the progression is
63
189
728
364
Solution
Let $a, a r, a r^2, a r^3, a r^4, a r^5$ be six terms of a G.P. where ' $a$ ' is first term and $\mathrm{r}$ is common ratio.
According to given conditions, we have $a r^3-a=5 \Rightarrow a\left(r^3-1\right)=52$ and $a+a r+a r^2=26$ $\Rightarrow a\left(1+r+r^2\right)=26$
To find: $a\left(1+r+r^2+r^3+r^4+r^5\right)$
Consider
$
\begin{aligned}
& a\left[1+r+r^2+r^3+r^4+r^5\right] \\
& =a\left[1+r+r^2+r^3\left(1+r+r^2\right)\right] \\
& =a\left[1+r+r^2\right]\left[1+r^3\right]
\end{aligned}
$
Divide (1) by (2), we get
$
\frac{r^3-1}{1+r+r^2}=2
$
we know $r^3-1=(r-1)\left(1+r+r^2\right)$
$
\begin{aligned}
& \therefore r-1=2 \Rightarrow r=3 \text { and } a=2 \\
& \therefore \quad a\left(1+r+r^2+r^3+r^4+r^5\right) \\
& \quad=a\left(1+r+r^2\right)\left(1+r^3\right) \\
& \quad=2(1+3+9)(1+27) \\
& \quad=26 \times 28=728
\end{aligned}
$