Question

Write the seventh term of the series $\frac{1}{\sqrt{2}},-2,\frac{8}{\sqrt{2}},\dots \dots$

Answer 1

$\frac{1}{\sqrt{2}},-2,\frac{8}{\sqrt{2}}\dots ..$ is a GP with $a=\frac{1}{\sqrt{2}}$.

Common ratio $=\frac{-2}{\frac{1}{\sqrt{2}}}=-2\sqrt{2}=\frac{\frac{8}{\sqrt{2}}}{-2}$

$T_7=a{r}^6=\frac{1}{\sqrt{2}}(-2\sqrt{2}{)}^6=\frac{1}{\sqrt{2}}(2^9)=256\sqrt{2}$.