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Q.Find the sum of the series .4+.44+.444+⋯.4 + .44 + .444 + \cdots to nn terms. OR Show that the ratio of the sum of first nn terms of a G.P to the sum of the terms from (n+1)th(n+1)^{th} to (2n)th(2n)^{th} terms is 1rn\dfrac{1}{r^n}.

Nagaland NbseNagaland Board of School Education (Class XI) 2025Subjective· 5mImportance★★★★★
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Pull out the common factor 4, rewrite each repunit-style term as 1−10−k1-10^{-k}, and sum the resulting GP.

Sn=0.4+0.44+0.444+⋯(n terms)=4(0.1+0.11+0.111+⋯ )S_n = 0.4+0.44+0.444+\cdots \text{(}n\text{ terms)} = 4(0.1+0.11+0.111+\cdots)

=49(0.9+0.99+0.999+⋯ )=49[(1−0.1)+(1−0.01)+(1−0.001)+⋯]= \frac{4}{9}(0.9+0.99+0.999+\cdots) = \frac49\Big[(1-0.1)+(1-0.01)+(1-0.001)+\cdots\Big]

=49[n−(0.1+0.01+0.001+⋯ to n terms)]= \frac49\left[n - (0.1+0.01+0.001+\cdots\text{ to }n\text{ terms})\right]

The bracketed geometric series has first term 0.10.1, ratio 0.10.1:

0.1+0.01+⋯=0.1(1−0.1n)1−0.1=19(1−10−n)0.1+0.01+\cdots = \frac{0.1(1-0.1^n)}{1-0.1} = \frac19\left(1-10^{-n}\right)

So

Sn=49[n−19(1−10−n)]=4n9−481(1−10−n)=481(9n−1+10−n)S_n = \frac49\left[n-\frac19\left(1-10^{-n}\right)\right] = \frac{4n}{9}-\frac{4}{81}\left(1-10^{-n}\right) = \frac{4}{81}\left(9n-1+10^{-n}\right)

Check (n=1n=1): 481(9−1+0.1)=481(8.1)=0.4\frac{4}{81}(9-1+0.1)=\frac{4}{81}(8.1)=0.4 ✓. Check (n=2n=2): 481(18−1+0.01)=481(17.01)=0.84=0.4+0.44\frac{4}{81}(18-1+0.01)=\frac{4}{81}(17.01)=0.84=0.4+0.44 ✓.


OR (alternative): Show the ratio of the sum of the first nn terms of a G.P. to the sum of the terms from (n+1)th(n+1)^{th} to (2n)th(2n)^{th} is 1rn\dfrac{1}{r^n}.

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