Question

If the sum and product of $20$ terms in G.P. are $16$ and $1024$ respectively, then the sum of its reciprocals is

Answer 1

$S=a+ar+a{r}^2+\cdots +a{r}^{n-1}\Rightarrow S=\frac{a(r^n-1)}{r-1}\Rightarrow \frac{a(r^{20}-1)}{r-1}=16\cdots \left(1\right)P=a.ar.a{r}^2...a^{n-1}\Rightarrow P=a^n.r^{1+2+3+\cdots +(n-1)}=a^n.r^{\frac{(n-1)n}{2}}\Rightarrow a^{20}{r}^{190}=1024$

Now, assuming
$R=\frac{1}{a}+\frac{1}{ar}+\cdots +\frac{1}{a{r}^{n-1}}\Rightarrow R=\frac{\frac{1}{a}({\left(\frac{1}{r}\right)}^{20}-1)}{\frac{1}{r}-1}$
$\Rightarrow R=\frac{a(1-r^{20})}{a^2{r}^{19}(1-r)}\Rightarrow R=\frac{16}{(1024{)}^{1/10}}\therefore R=\frac{16}{2}=8$