# Question #1df71

Mar 20, 2017

Given that first term of geometric series $a = - 11$,
the common ratio $r = - 4$,
the last or 8 th term ${t}_{8} = a {r}^{7} = 180224$

Now

${S}_{8} = a + a r + a {r}^{2} + a {r}^{3} + \ldots \ldots . a {r}^{7} - - - - \left(1\right)$

$r {S}_{8} = a r + a {r}^{2} + a {r}^{3} + \ldots \ldots + a {r}^{7} + a {r}^{8} - - - - \left(2\right)$

subtracting (2) from (1) we get

$\left(1 - r\right) {S}_{8} = a - a {r}^{8} = a - r \times a {r}^{7}$

$\implies {S}_{8} = \frac{a - r \times {t}_{8}}{1 - r} = \frac{- 11 - \left(- 4\right) \times 180224}{1 - \left(- 4\right)}$

$= \frac{- 11 + 720896}{5} = 144177$