The difference between the fourth and the first term of a geometric progression (G.P) is $52$ and the sum of the first three terms is $26$. Calculate the sum of the first six terms of the progression.
Let the first term of the G.P be $= t_1 = a$
and the common ratio be $= r$
$n^{th}$ term of G.P $= t_n = a \times r^{n - 1}$
$\therefore \;$ $4^{th}$ term of G.P $= t_4 = a \cdot r^3$
Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(r^n - 1\right)}{r - 1}$
Given: $\;$ $t_4 - t_1 = 52$
i.e. $\;$ $a \cdot r^3 - a = 52$
i.e. $\;$ $a \left(r^3 - 1\right) = 52$ $\;\;\; \cdots \; (1)$
And: $\;$ $S_3 = 26$
i.e. $\;$ $\dfrac{a \left(r^3 - 1\right)}{r - 1} = 26$ $\;\;\; \cdots \; (2)$
In view of equation $(1)$, equation $(2)$ becomes
$\dfrac{52}{r - 1} = 26$
i.e. $\;$ $r - 1 = 2$ $\implies$ $r = 3$
Substituting the value of $r$ in equation $(1)$ gives
$a \left(3^3 - 1\right) = 52$ $\implies$ $a = \dfrac{52}{26} = 2$
$\therefore \;$ Sum of the first six terms of G.P
$= S_6 = \dfrac{a \left(r^6 - 1\right)}{r - 1} = \dfrac{2 \times \left(3^6 - 1\right)}{3 - 1} = 728$