Let a_1,a_2,… ,a_21 be an A.P. such that Σ _n=1^20 1 a_n a_n+1= 4 9. If the sum of this A.P. is 189, then a_6 a_16 is equal to :

$\displaystyle \sum _{n=1}^{20}\frac{1}{{a}_{n}{a}_{n+1}}=\frac{4}{9}$n=1∑20​an​an+1​1​=94​ $\text{ [let common difference of }\mathrm{AP}\mathrm{is}d={a}_{n+1}-{a}_{n}]$ [let common difference of APisd=an+1​−an​]

$$=\frac{1}{d}\sum _{n=1}^{20}(\frac{1}{{a}_{n}}-\frac{1}{{a}_{n+1}})\Rightarrow \frac{1}{d}[(\frac{1}{{a}_{1}}-\frac{1}{{a}_{2}})+(\frac{1}{{a}_{2}}-\frac{1}{{a}_{3}})+\ldots .(\frac{1}{{a}_{20}}-\frac{1}{{a}_{21}})]=\frac{4}{9}$$=d1​n=1∑20​(an​1​−an+1​1​)⇒d1​[(a1​1​−a2​1​)+(a2​1​−a3​1​)+….(a20​1​−a21​1​)]=94​

$$\Rightarrow \frac{1}{d}(\frac{1}{{a}_{1}}-\frac{1}{{a}_{21}})=\frac{4}{9}$$⇒d1​(a1​1​−a21​1​)=94​

$$\Rightarrow \frac{1}{d}(\frac{{a}_{21}-{a}_{1}}{{a}_{1}\cdot {a}_{21}})=\frac{4}{9}\Rightarrow \frac{20d}{d({a}_{1}{a}_{21})}=\frac{4}{9}\Rightarrow {a}_{1}{a}_{21}=45$$⇒d1​(a1​⋅a21​a21​−a1​​)=94​⇒d(a1​a21​)20d​=94​⇒a1​a21​=45

$${a}_{1}({a}_{1}+20d)=45...(1)$$a1​(a1​+20d)=45...(1)

Now sum of first $21$21 terms$=\frac{21}{2}(2{a}_{1}+20d)=189\Rightarrow {a}_{1}+10d=9...(2)$=221​(2a1​+20d)=189⇒a1​+10d=9...(2)

by using equation $(1)$(1) and $(2)$(2) we get ${a}_{1}=3,d=\frac{3}{5}$a1​=3,d=53​ otherwise ${a}_{1}=15,d=-\frac{3}{5}$a1​=15,d=−53​

So, ${a}_{6}{a}_{16}=({a}_{1}+5d)({a}_{1+}+15d)=72$a6​a16​=(a1​+5d)(a1+​+15d)=72