The 2nd and 5th terms of a geometric series are $-\dfrac{1}{2} \text{ and } \dfrac{1}{16}$ respectively. Find the sum of the series upto 8 terms.

Given,

$a$2 = -\dfrac{1}{2} and $a5 = \dfrac{1}{16}$

By formula, $a_n = ar^{n - 1}$ we get,

$\Rightarrow a$2 = ar \text{ and } a5 = ar^4

Dividing $a$5 by $a2$,

$\Rightarrow \dfrac{ar^4}{ar} = \dfrac{\dfrac{1}{16}}{-\dfrac{1}{2}} \\[1em] \Rightarrow r^3 = -\dfrac{1}{8} \\[1em] \Rightarrow r^3 = \Big(-\dfrac{1}{2}\Big)^3 \\[1em] \therefore r = -\dfrac{1}{2}.$

Since, $a_2 = ar$ or,

$\Rightarrow -\dfrac{1}{2} = a\Big(-\dfrac{1}{2}\Big) \\[1em] \Rightarrow a = 1.$

By formula,

$S$n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S8 = \dfrac{1\Big[\Big(-\dfrac{1}{2}\Big)^8 - 1\Big]}{-\dfrac{1}{2} - 1} \\[1em] = \dfrac{\dfrac{1}{256} - 1}{\dfrac{-1 - 2}{2}} \\[1em] = \dfrac{\dfrac{1 - 256}{256}}{\dfrac{-3}{2}} \\[1em] = \dfrac{-255 \times 2}{256 \times (-3)} \\[1em] = \dfrac{-510}{-768}

Dividing numerator and denominator by 6,

$= \dfrac{85}{128}.$

Hence, the sum of the series upto 8 terms is $\dfrac{85}{128}$.