If 2^{10}+2^{9} · 3^{1}+28 · 3^{2}+ldots+2 · 3^{9}+3^{10}=S -211 then S is equal to

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8. Sequences and Series

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If $2^{10}+2^{9} \cdot 3^{1}+28 \cdot 3^{2}+\ldots+2 \cdot 3^{9}+3^{10}=S -211$ then $S$ is equal to

A

$\frac{3^{11}}{2}+2^{10}$

B

$3^{11}-2^{12}$

C

$3^{11}$

D

$2 \cdot 3^{11}$

(JEE MAIN-2020)

Solution

$a =2^{10} ; r =\frac{3}{2} ; n =11( G \cdot P )$

$S^{\prime}=\left(2^{10}\right) \frac{\left(\left(\frac{3}{2}\right)^{11}-1\right)}{\frac{3}{2}-1}=2^{11}\left(\frac{3^{11}}{2^{11}}-1\right)$

$S^{\prime}=3^{11}-2^{11}= S -2^{11}( Given )$

$\therefore S =3^{11}$

Standard 11

Mathematics

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