The sum of the first three terms of a geometric progression (G.P) is $31$ and the sum of the first and the third term is $26$. Find the seventh term of the progression.
Let the given G.P be $b_1, \; b_2, \; b_3, \; b_4, \cdots$
Let the first term of the G.P be $= b_1 = \dfrac{a}{r}$;
second term $= b_2 = a$
and the third term $= b_3 = a \cdot r$
where $r$ is the common ratio of the G.P
$n^{th}$ term of the progression $= b_n = b_1 \times r^{n - 1}$
$\therefore \;$ Seventh term of G.P $= b_7 = b_1 \times r^6 = \dfrac{a}{r} \times r^6 = a\cdot r^5$
Given: $\;$ $b_1 + b_2 + b_3 = 31$
i.e. $\;$ $\dfrac{a}{r} + a + a \cdot r = 31$
i.e. $\;$ $\dfrac{a}{r} + a \cdot r = 31 - a$ $\;\;\; \cdots \; (1)$
And: $\;$ $b_1 + b_3 = 26$
i.e. $\;$ $\dfrac{a}{r} + a \cdot r = 26$ $\;\;\; \cdots \; (2)$
i.e. $\;$ $31 - a = 26$ $\;\;\;$ [in view of equation $(1)$]
$\implies$ $a = 5$
Substituting $a = 5$ in equation $(2)$ gives
$5 \left(\dfrac{1}{r} + r\right) = 26$
i.e. $\;$ $5 r^2 - 26 r + 5 = 0$
i.e. $\;$ $\left(5r - 1\right) \left(r - 5\right) = 0$
$\implies$ $r = \dfrac{1}{5}$ $\;$ or $\;$ $r = 5$
$\therefore \;$ When $a = 5$, $r = \dfrac{1}{5}$, the $7^{th}$ term of G.P $= 5 \times \left(\dfrac{1}{5}\right)^5 = \dfrac{1}{625}$
and when $a = 5$, $r = 5$, the $7^{th}$ term of G.P $= 5 \times 5^5 = 15625$