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in Arithmetic Progression by (15 points)
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 AP is 189 , then \( a_{6} a_{16} \) is equal to (1) 57 (2) 72 (3) 48 [JEE-Main Sept. 01, 2021 Shift-II (4) 36

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\(a_1 + a_2 + .... + a_{21} = 189\)

⇒ \(\frac{21}2(a_1 + a_{21}) = 189\)

⇒ \(a_1 + a_{21} = 189 \times \frac2{21} = 9 \times 2 = 18\)

Now,

\(\sum\limits^{20}_{n=1} \frac1{a_na_{n+ 1}} = \frac 49\)

⇒ \(\frac 1d\sum\limits^{20}_{n=1} \frac{a_{n + 1} - a_n}{a_na_{n+ 1}} = \frac 49\)

⇒ \(\frac1d \left(\sum\limits^{20}_{n = 1} \left(\frac1{a_n} - \frac1{a_{n + 1}}\right)\right) = \frac49\)

⇒ \(\frac1d \left(\left(\frac1{a_1} - \frac1{a_2}\right) + \left(\frac1{a_2} - \frac1{a_3}\right) + .....+ \left(\frac1{a_{20}} - \frac1{a_{21}}\right)\right)= \frac49\)

⇒ \(\frac1d\left(\frac1 {a_1} - \frac1{a_{21}}\right) = \frac49\)

⇒ \(\frac{a_{21} - a_1}{a_1a_{21}} = \frac49 d\)

⇒ \(\frac{(a_{21} - a_{20}) + (a_{20} - a_{19}) + ....+(a_2 - a_1)}{a_1 a_{21}} = \frac49 d\)

⇒ \(\frac{20d}{a_1a_{21}} = \frac49d\)      \((\because a_{n+1} - a_n = d)\)

⇒ \(a_1a_{21} = \frac94 \times 20 = 45\)

Hence, we have 

\(a_1 + a_{21} = 18\)     ...(1)

\(a_1 a_{21} = 45\)

\((a_1 - a_{21})^2 = (a_1 + a_{21})^2 - 4a_1 a_{21} \)

\(= 18^2 - 4 \times 45\)

\(= 324 - 180\)

\(= 144 \)

\(= 12^2\)

⇒ \(a_1-a_{21} = 12\)    ....(2)

By adding (1) & (2), we get 

\(2a_1 = 30\)

⇒ \(a_1 = 15\)

\(\therefore a_{21} = 18 - a_1 = 18 - 15 = 3\)

\(\because a_{21} - a_1 = 20d\)

⇒ \(d = \frac{a_{21} - a_1}{20} = \frac{-12}{20} = \frac{-3}{5}\)

\(\therefore a_6 = a + 5d = 15 - \frac35 \times 5 = 15 - 3 = 12\)

\(a_{16} = a + 15 d = 15 + 15 \times \frac{-3}5= 15 - 9 = 6\)

\(\therefore a_6a_{16} = 12 \times 6 = 72\)

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