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Q.Find the sum of nn terms of the sequence 8,88,888,8888,…8, 88, 888, 8888, \ldots OR If the A.M. and G.M. of two positive numbers "a" and "b" are 1010 and 88 respectively, then find the numbers.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2026Subjective· 3mImportance★★★★★
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Factoring out 8 and rewriting each term as 19(10k−1)\tfrac19(10^k-1) converts the series into a geometric sum, giving Sn=881(10n+1−9n−10)S_n=\dfrac{8}{81}(10^{n+1}-9n-10).

The series is Sn=8+88+888+⋯S_n = 8+88+888+\cdots to nn terms.

Factor 8 from every term: Sn=8(1+11+111+⋯to n terms)S_n = 8(1+11+111+\cdots\text{to }n\text{ terms}).

Each term 1,11,111,…1,11,111,\ldots can be written as 10k−19\dfrac{10^k-1}{9} for the kthk^{\text{th}} term (e.g. 11=102−1911=\tfrac{10^2-1}{9}).

So Sn=89∑k=1n(10k−1)=89[∑k=1n10k−n]S_n = \dfrac{8}{9}\sum_{k=1}^{n}(10^k-1) = \dfrac{8}{9}\left[\sum_{k=1}^n 10^k - n\right].

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