The sequence $\left\{b_n\right\}$ is a geometric progression (G.P) with $\dfrac{b_4}{b_6} = \dfrac{1}{4}$ and $b_2 + b_5 = 216$. Find $b_1$.
Let the first term of the G.P be $= b_1$
and the common ratio be $= r$
$n^{th}$ term of G.P $= b_n = b_1 \cdot r^{n-1}$
$\therefore \;$ Second term of G.P $= b_2 = b_1 \cdot r$
Fourth term of G.P $= b_4 = b_1 \cdot r^3$
Fifth term of G.P $= b_5 = b_1 \cdot r^4$
Sixth term of G.P $= b_6 = b_1 \cdot r^5$
Given: $\;$ $\dfrac{b_4}{b_6} = \dfrac{1}{4}$
i.e. $\;$ $\dfrac{b_1 \cdot r^3}{b_1 \cdot r^5} = \dfrac{1}{4}$
i.e. $\;$ $\dfrac{1}{r^2} = \dfrac{1}{4}$ $\implies$ $r^2 = 4$ $\implies$ $r = \pm 2$
And: $\;$ $b_2 + b_5 = 216$
i.e. $\;$ $b_1 \cdot r + b_1 \cdot r^4 = 216$
i.e. $\;$ $b_1 \left(r + r^4\right) = 216$ $\implies$ $b_1 = \dfrac{216}{r + r^4}$
$\therefore \;$ When $\;$ $r = + 2$, $\;$ $b_1 = \dfrac{216}{2 + 2^4} = \dfrac{216}{18} = 12$
When $\;$ $r = - 2$, $\;$ $b_1 = \dfrac{216}{- 2 + \left(-2\right)^4} = \dfrac{216}{14} = \dfrac{108}{7}$