JEE Main 2021 (Online) 1st September Evening ShiftQuestion 1
Let $a_1,a_2,\ldots,a_{21}$ be an A.P. such that $\sum_{n=1}^{20}\frac{1}{a_na_{n+1}}=\frac49$. If the sum of this A.P. is $189$, then $a_6a_{16}$ is equal to :
Sequence and Series · Arithmetic Progression · Telescoping ($V_n$ Method)
Solution
Formulas used. (1) $\sum_{n=1}^{N}\frac1{a_na_{n+1}}=\frac1d\left(\frac1{a_1}-\frac1{a_{N+1}}\right)=\frac{N}{a_1a_{N+1}}$(2) $S_n=\frac n2\,(a_1+a_n)$Step 1 - Product of end terms. With $N=20$: $$\frac{20}{a_1a_{21}}=\frac49\ \Longrightarrow\ a_1a_{21}=45 .$$ Step 2 - Sum of end terms. $$\frac{21}{2}\left(a_1+a_{21}\right)=189\ \Longrightarrow\ a_1+a_{21}=18 .$$ Step 3 - End terms. $a_1,a_{21}$ are roots of $t^2-18t+45=0$: $\{3,15\}$. Take $a_1=3,\ a_{21}=15$; then $d=\dfrac{12}{20}=\dfrac35$. (The other order gives the same A.P. reversed.) Step 4. $$a_6=3+5\cdot\frac35=6,\qquad a_{16}=3+15\cdot\frac35=12,\qquad a_6a_{16}=72 .$$ Answer: (D)