Mathematics - Sequence & Series Question with Solution | TestHub

Let $\mathrm{GP}: \mathrm{a}, \mathrm{ar}, {\mathrm{ar}}^2, {\mathrm{ar}}^3,...$

Now, given: $2\mathrm{ar}={\mathrm{ar}}^2+{\mathrm{ar}}^3$

$\Rightarrow 2=\mathrm{r}+{\mathrm{r}}^2$

$\Rightarrow {\mathrm{r}}^2+\mathrm{r}-2=0$

$\Rightarrow \left(\mathrm{r}+2\right)\left(\mathrm{r}-1\right)=0$

$\Rightarrow \mathrm{r}=-2, 1 \left[\text{Since} \mathrm{GP} \text{is not constant so }r=1\text{ is rejected}\right]$

$\Rightarrow \mathrm{r}=-2$

$\Rightarrow {\mathrm{S}}_{20}-{\mathrm{S}}_{18}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{20}\right)}{1-\mathrm{r}}-\frac{\mathrm{a}\left(1-{\mathrm{r}}^{18}\right)}{1-\mathrm{r}}$

$\Rightarrow {\mathrm{S}}_{20}-{\mathrm{S}}_{18}=\frac{\mathrm{a}\left[{\mathrm{r}}^{18}-{\mathrm{r}}^{20}\right]}{1-\mathrm{r}}$

$\Rightarrow {\mathrm{S}}_{20}-{\mathrm{S}}_{18}=\frac{\left[2^{18}-2^{20}\right]}{8\times 3}$

$\Rightarrow {\mathrm{S}}_{20}-{\mathrm{S}}_{18}=\frac{2^{18}\left(1-4\right)}{24}$

$\Rightarrow {\mathrm{S}}_{20}-{\mathrm{S}}_{18}=-2^{15}$