Question

Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an AP such that $\displaystyle\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. 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)

$\displaystyle\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\displaystyle\sum_{n=1}^{20} \frac{1}{a_{n}\left(a_{n}+d\right)}$
$=\frac{1}{d} \displaystyle\sum_{n=1}^{20}\left(\frac{1}{a_{n}}-\frac{1}{a_{n}+d}\right)$
$\Rightarrow \frac{1}{d}\left(\frac{1}{a_{1}}-\frac{1}{a_{21}}\right)=\frac{4}{9}$ (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}+20 d-a_{1}}{a_{1} a_{2}}\right)=\frac{4}{9} $
$\Rightarrow a_{1} a_{2}=45$ ...(1)
Now sum of first $21$ terms $=\frac{21}{2}\left(2 a _{1}+20 d \right)=189$
$\Rightarrow a_{1}+10 d =9$ ...(2)
For equation (1) & (2) we get
$a _{1}=3\, \& \,d =\frac{3}{5}$
$a _{1}=15 \,\& \,d =-\frac{3}{5}$
So, $a _{6} \cdot a _{16}=\left( a _{1}+5 d \right)\left( a _{1}+15 d \right)$
$\Rightarrow a _{6} a _{16}=72$