C-13-Q103mediumsingle_mcqThe general term of 128+64+32+⋯128 + 64 + 32 + \cdots128+64+32+⋯ is —a28−n2^{8 - n}28−nb2n−82^{n - 8}2n−8c128⋅2n−1128 \cdot 2^{n - 1}128⋅2n−1d128n\dfrac{128}{n}n128ব্যাখ্যাWith a=128=27a=128=2^{7}a=128=27 and r=12r=\tfrac12r=21, the general term is 27⋅(2−1)n−1=27−(n−1)=28−n2^{7}\cdot\left(2^{-1}\right)^{n-1}=2^{7-(n-1)}=2^{8-n}27⋅(2−1)n−1=27−(n−1)=28−n. This correctly gives 27=1282^{7}=12827=128 at n=1n=1n=1.