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(A) Given the sum of an infinite $G.P.$ is $S_{\infty} = \frac{a}{1-r} = 5$,so $a = 5(1-r)$.The sum of the first five terms is $S_5 = \frac{a(1-r^5)}{

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$21 a_{11}$

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(A) Given the sum of an infinite $G.P.$ is $S_{\infty} = \frac{a}{1-r} = 5$,so $a = 5(1-r)$.The sum of the first five terms is $S_5 = \frac{a(1-r^5)}{1-r} = 5(1-r^5) = \frac{98}{25}$.$1-r^5 = \frac{98}{125} \implies r^5 = 1 - \frac{98}{125} = \frac{27}{125} = (\frac{3}{5})^5$,so $r = \frac{3}{5}$.Then $a = 5(1 - \frac{3}{5}) = 5(\frac{2}{5}) = 2$.For the $A.P.$,the first term $A = 10ar = 10(2)(\frac{3}{5}) = 12$ and the common difference $D = 10ar^2 = 10(2)(\frac{9}{25}) = \frac{36}{5} = 7.2$.The sum of the first $21$ terms is $S_{21} = \frac{21}{2} [2A + (21-1)D] = \frac{21}{2} [2(12) + 20(7.2)] = \frac{21}{2} [24 + 144] = \frac{21}{2} [168] = 21 	imes 84 = 1764$.Now,calculate $a_{11}$ of the $A.P.$: $a_{11} = A + 10D = 12 + 10(7.2) = 12 + 72 = 84$.Thus,$S_{21} = 21 	imes 84 = 21 a_{11}$.

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