Let a_{1}, a_{2}, ldots, a_{21} be an A.P. su | vedclass

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(B) Given $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}} = \frac{4}{9}$.Since $a_{n+1} = a_{n} + d$,we have $\frac{1}{a_{n} a_{n+1}} = \frac{1}{d} \left( \f

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$72$

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(B) Given $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}} = \frac{4}{9}$.Since $a_{n+1} = a_{n} + d$,we have $\frac{1}{a_{n} a_{n+1}} = \frac{1}{d} \left( \frac{1}{a_{n}} - \frac{1}{a_{n+1}} ight)$.Thus,$\frac{1}{d} \sum_{n=1}^{20} \left( \frac{1}{a_{n}} - \frac{1}{a_{n+1}} ight) = \frac{1}{d} \left( \frac{1}{a_{1}} - \frac{1}{a_{21}} ight) = \frac{4}{9}$.$\frac{1}{d} \left( \frac{a_{21} - a_{1}}{a_{1} a_{21}} ight) = \frac{1}{d} \left( \frac{20d}{a_{1} a_{21}} ight) = \frac{20}{a_{1} a_{21}} = \frac{4}{9} \implies a_{1} a_{21} = 45$.Sum of $21$ terms $S_{21} = \frac{21}{2} (a_{1} + a_{21}) = 189 \implies a_{1} + a_{21} = 18$.We have $a_{1} + a_{21} = 18$ and $a_{1} a_{21} = 45$. The roots of $x^{2} - 18x + 45 = 0$ are $a_{1}, a_{21}$.$(x - 15)(x - 3) = 0 \implies \{a_{1}, a_{21}\} = \{3, 15\}$.Case $1$: $a_{1} = 3, a_{21} = 15 \implies 3 + 20d = 15 \implies d = 0.6$.Case $2$: $a_{1} = 15, a_{21} = 3 \implies 15 + 20d = 3 \implies d = -0.6$.$a_{6} a_{16} = (a_{1} + 5d)(a_{1} + 15d)$.For Case $1$: $(3 + 5(0.6))(3 + 15(0.6)) = (3 + 3)(3 + 9) = 6 	imes 12 = 72$.For Case $2$: $(15 + 5(-0.6))(15 + 15(-0.6)) = (15 - 3)(15 - 9) = 12 	imes 6 = 72$.Thus,$a_{6} a_{16} = 72$.

Let $a_{1}, a_{2}, \ldots, a_{21}$ 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 $A.P.$ is $189$,then $a_{6} a_{16}$ is equal to:

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