Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an AP such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. If the sum of this $A P$ is 189 , then a a is equal to :
Correct Option: , 2
$\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\sum_{n=1}^{20} \frac{1}{a_{n}\left(a_{n}+d\right)}$
$=\frac{1}{d} \sum_{n=1}^{20}\left(\frac{1}{a_{n}}-\frac{1}{a_{n}+d}\right)$
$\Rightarrow \frac{1}{d}\left(\frac{1}{a_{1}}-\frac{1}{a_{21}}\right)=\frac{4}{9}$ (Given)
$\Rightarrow \frac{1}{d}\left(\frac{a_{21}-a_{1}}{a_{1} a_{21}}\right)=\frac{4}{9}$
$\Rightarrow \frac{1}{d}\left(\frac{a_{1}+20 d-a_{1}}{a_{1} a_{2}}\right)=\frac{4}{9} \Rightarrow a_{1} a_{2}=45$
Now sum of first 21 terms $=\frac{21}{2}\left(2 \mathrm{a}_{1}+20 \mathrm{~d}\right)=189$
$\Rightarrow a_{1}+10 d=9 \ldots$(2)
For equation (1) \& (2) we get
$a_{1}=3 \& d=\frac{3}{5}$
OR
$a_{1}=15 \& d=-\frac{3}{5}$
So, $a_{6} \cdot a_{16}=\left(a_{1}+5 d\right)\left(a_{1}+15 d\right)$
$\Rightarrow a_{6} a_{16}=72$
Option (2)
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