Let the sum of an infinite G.P., whose first term is a and common ratio is r, be 5.

Question

Let the sum of an infinite G.P., whose first term is $a$ and common ratio is $r$, be $5$. Let the sum of its first five terms be $\dfrac{98}{25}$. Then the sum of the first $21$ terms of an A.P., whose first term is $10ar$, $n$th term is $a_n$, and common difference is $10ar^2$, is equal to:

Accepted Answer

$\dfrac{a}{1-r}=5$ and $\dfrac{a(1-r^5)}{1-r}=\dfrac{98}{25} \Rightarrow 5(1-r^5)=\dfrac{98}{25} \Rightarrow r^5=\dfrac{27}{125} \Rightarrow r=\dfrac35$, giving $a=2$.

For any A.P. with an odd number of terms, the sum equals (number of terms) $\times$ (middle term): with $21$ terms, the middle term is $a_{11}$, so

$$S_{21} = 21\,a_{11}.$$

Solutions · 1