The $2^{nd}$ and $5^{th}$ terms of a G.P are $\dfrac{1}{2}$ and $\dfrac{1}{16}$ respectively. Find the sum of the first eight terms of the G.P. Give the answer as a fraction.
Let the first term of G.P be $= a$
Let the common ratio of G.P be $= r$
$n^{th}$ term of G.P $= t_n = ar^{n-1}$
Given: $\;$ $2^{nd}$ term of G.P $= t_2 = ar = \dfrac{1}{2}$ $\;\;\; \cdots \; (1)$
$5^{th}$ term of G.P $= t_5 = ar^4 = \dfrac{1}{16}$ $\;\;\; \cdots \; (2)$
Dividing equations $(1)$ and $(2)$ we have,
$\dfrac{ar^4}{ar} = \dfrac{1/16}{1/2}$
i.e. $\;$ $r^3 = \dfrac{1}{8} = \left(\dfrac{1}{2}\right)^3$
$\implies$ $r = \dfrac{1}{2}$
Substituting the value of $r$ in equation $(1)$ gives,
$a \times \dfrac{1}{2} = \dfrac{1}{2}$ $\implies$ $a = 1$
Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(1 - r^n\right)}{1 - r}$ $\;$ when $\;$ $r < 1$
$\therefore \;$ Sum of first $8$ terms of G.P $= S_8 = \dfrac{1 \left[1 - \left(\dfrac{1}{2}\right)^8\right]}{1 - \dfrac{1}{2}}$
i.e. $\;$ $S_8 = \dfrac{1 - \dfrac{1}{256}}{\dfrac{1}{2}} = \dfrac{255 / 256}{1 / 2} = \dfrac{255}{128}$