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Q.Find the sum to nn terms of the sequence 8,88,888,…8, 88, 888, \ldots OR Find the two numbers between 33 and 8181 so that the resulting sequence is a Geometric Progression.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2023Subjective· 4mImportance★★★★★
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The sum to nn terms of 8,88,888,…8,88,888,\ldots is Sn=881(10n+1−9n−10)S_n = \dfrac{8}{81}(10^{n+1}-9n-10).

Write Sn=8+88+888+⋯S_n = 8 + 88 + 888 + \cdots to nn terms =8(1+11+111+⋯ to n terms)= 8(1 + 11 + 111 + \cdots \text{ to } n \text{ terms}).

Each term in the bracket can be written using 99: 1=10−191 = \dfrac{10-1}{9}, 11=100−1911 = \dfrac{100-1}{9}, 111=1000−19111=\dfrac{1000-1}{9}, and so on.

So Sn=89[(10−1)+(102−1)+(103−1)+⋯+(10n−1)]S_n = \dfrac{8}{9}\left[(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^n-1)\right]

=89[(10+102+⋯+10n)−n]= \dfrac{8}{9}\left[(10+10^2+\cdots+10^n) - n\right]

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