Question

The second term of a geometric series is $3$ and the common ratio is $\frac{4}{5}$. Find the sum of first $23$ consecutive terms in the given geometric series

Answer 1

It is given that the second term term of the geometric series is $T_2=3$, the common ratio is $r=\frac{4}{5}<1$.

We know that the general term of an geometric progression with first term $a$ and common ratio $r$ is $T_n=a{r}^{n-1}$, therefore,

$T_2=a{\left(\frac{4}{5}\right)}^{2-1}\Rightarrow 3=\frac{4a}{5}\Rightarrow a=\frac{3\times 5}{4}=\frac{15}{4}$

We also know that the sum of an geometric series with first term $a$ and common ratio $r$ is $S_n=\frac{a(1-r^n)}{1-r}$ if $r<1$

Now, to find the sum of first $23$ consecutive terms, substitute $a=\frac{15}{4},r=\frac{4}{5}$ and $n=23$ in $S_n=\frac{a(1-r^n)}{1-r}$ as follows:

$S_{23}=\frac{\frac{15}{4}[1-{\left(\frac{4}{5}\right)}^{23}]}{1-\frac{4}{5}}=\frac{\frac{15}{4}[1-{\left(\frac{4}{5}\right)}^{23}]}{\frac{5-4}{5}}=\frac{\frac{15}{4}[1-{\left(\frac{4}{5}\right)}^{23}]}{\frac{1}{5}}=\frac{15}{4}\times 5[1-{\left(\frac{4}{5}\right)}^{23}]=\frac{75}{4}[1-{\left(\frac{4}{5}\right)}^{23}]$

Hence the sum is $\frac{75}{4}[1-{\left(\frac{4}{5}\right)}^{23}]$.