JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a_{1}, a_{2}, \ldots \ldots, a_{21}\) be an \(A.P.\) such that \(\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}\). If the sum of this AP is \(189,\) then \(a_{6} \mathrm{a}_{16}\) is equal to :
- A \(57\)
- B \(72\)
- C \(48\)
- D \(36\)
Answer & Solution
Correct Answer
(B) \(72\)
Step-by-step Solution
Detailed explanation
\(\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\sum_{n=1}^{20} \frac{1}{a_{n}\left(a_{n}+d\right)}\) \(=\frac{1}{d} \sum_{n=1}^{20}\left(\frac{1}{a_{n}}-\frac{1}{a_{n}+d}\right)\) \(\Rightarrow \frac{1}{d}\left(\frac{1}{a_{1}}-\frac{1}{a_{21}}\right)=\frac{4}{9} \text { (Given) }\)…
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