Let the sum of an infinite G . P . , whose first term is a and the...

For the infinite G.P.

${S}_{\infty} = \dfrac{a}{1 - r} = 5$

Also $a + ar + a{r}^{2} + a{r}^{3} + a{r}^{4} = 5$

$⇒ a\left(\dfrac{1 - {r}^{5}}{1 - r}\right) = \dfrac{98}{25}$

i.e. $1 - {r}^{5} = \dfrac{98}{125} ⇒ {r}^{5} = \dfrac{{3}^{3}}{{5}^{3}} ⇒ r = \left(\dfrac{3}{5}\right)\dfrac{3}{5}$

Now ${S}_{21} = \dfrac{21}{2}\left[20ar + \left(21 - 1\right)10a{r}^{2}\right]$

$= 21\left[10ar + 100a{r}^{2}\right]$

$= 21 \cdot {a}_{11}$