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Q.Find the sum to nn terms of the sequence 8,88,888,8888,…8, 88, 888, 8888, \ldots OR The sum of the first three terms of a G.P. is 3910\dfrac{39}{10} and their product is 11. Find the common ratio and the terms.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2025Subjective· 4mImportance★★★★★
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The sum of nn terms of 8,88,888,…8, 88, 888, \ldots is Sn=881(10n+1−9n−10)S_n = \dfrac{8}{81}\left(10^{n+1} - 9n - 10\right).

The general term can be written as: ak=811…1⏟k ones=89(10k−1)a_k = 8\underbrace{11\ldots1}_{k\text{ ones}} = \dfrac{8}{9}(10^k - 1) for k=1,2,…,nk = 1, 2, \ldots, n (check: k=1k=1: 89(9)=8\dfrac{8}{9}(9) = 8 ✓; k=2k=2: 89(99)=88\dfrac{8}{9}(99)=88 ✓).

Sum to nn terms:

Sn=∑k=1n89(10k−1)=89(∑k=1n10k−n)S_n = \displaystyle\sum_{k=1}^{n} \dfrac{8}{9}(10^k - 1) = \dfrac{8}{9}\left(\sum_{k=1}^{n}10^k - n\right).

The geometric sum ∑k=1n10k=10(10n−1)9=10n+1−109\displaystyle\sum_{k=1}^{n}10^k = \dfrac{10(10^n-1)}{9} = \dfrac{10^{n+1}-10}{9}.

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