Question

Let $a_1,a_2,\dots \dots ,a_{21}$ be an AP such that $\sum _{n=1}^{20}\frac{1}{a_n{a}_{n+1}}=\frac{4}{9}$. If the sum of this $AP$ is 189, then $a_6{a}_{16}$ is equal to:

• A. 57
• B. 72
• C. 48
• D. 36

Answer 1

Answer: (B)

$\sum _{n=1}^{20}\frac{1}{a_n{a}_{n+1}}=\sum _{n=1}^{20}\frac{1}{a_n\left(a_n+d\right)}$
$=\frac{1}{d}\sum _{n=1}^{20}\left(\frac{1}{a_n}-\frac{1}{a_n+d}\right)$
$\Rightarrow \frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{21}}\right)=\frac{4}{9}$ (Given)
$\Rightarrow \frac{1}{d}\left(\frac{a_{21}-a_1}{a_1{a}_{21}}\right)=\frac{4}{9}$
$\Rightarrow \frac{1}{d}\left(\frac{a_1+20d-a_1}{a_1{a}_2}\right)=\frac{4}{9}$
$\Rightarrow a_1{a}_2=45$ ...(1)
Now sum of first $21$ terms $=\frac{21}{2}\left(2a_1+20d\right)=189$
$\Rightarrow a_1+10d=9$ ...(2)
For equation (1) & (2) we get
$a_1=3\&d=\frac{3}{5}$
$a_1=15\&d=-\frac{3}{5}$
So, $a_6\cdot a_{16}=\left(a_1+5d\right)\left(a_1+15d\right)$
$\Rightarrow a_6{a}_{16}=72$