QJEEM21Maa0b4: JEE Main 2021 Mathematics - Class 11 JEEM21Maa0b4 Solution: Let a 1 , a 2 , ..........., a 21 be an AP such that . If the…

Question

Verified Solution • 2025–26 Session moderate

Let a₁, a₂, ..........., a₂₁ be an AP such that

\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = {4 \over 9}}

. If the sum of this AP is 189, then a₆a₁₆ is equal to :

Accepted Answer

\sum\limits_{n = 1}^{20} {{1 \over {{a_n}{a_{n + 1}}}} = \sum\limits_{n = 1}^{20} {{1 \over {{a_n}({a_n} + d)}}} }

= {1 \over d}\sum\limits_{n = 1}^{20} {\left( {{1 \over {{a_n}}} - {1 \over {{a_n} + d}}} \right)}

\Rightarrow {1 \over d}\left( {{1 \over {{a_1}}} - {1 \over {{a_{21}}}}} \right) = {4 \over 9}

(Given)

\Rightarrow {1 \over d}\left( {{{{a_{21}} - {a_1}} \over {{a_1}{a_{21}}}}} \right) = {4 \over 9}

\Rightarrow {1 \over d}\left( {{{{a_1} + 20d - {a_1}} \over {{a_1}{a_2}}}} \right) = {4 \over 9} \Rightarrow {a_1}{a_2} = 45

.... (1)

Now sum of first 21 terms =

{{21} \over 2}(2{a_1} + 20d) = 189

\Rightarrow

a₁ + 10d = 9 ..... (2)

For equation (1) & (2) we get

a₁ = 3 & d =

{3 \over 5}

or a₁ = 15 & d =

- {3 \over 5}

So, a₆ . a₁₆ = (a₁ + 5d) (a₁ + 15d)

\Rightarrow

a₆a₁₆ = 72

Option (b)

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