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Q.Find the sum of the following series up to nn terms: 5+55+555+5555+…5+55+555+5555+\ldots OR If the A.M. and G.M. of two positive numbers aa and bb are 1010 and 88 respectively, find the numbers.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2022Subjective· 4mImportance★★★★★
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The sum of 5+55+555+…5+55+555+\ldots to nn terms is Sn=581(10n+1−9n−10)S_n = \dfrac{5}{81}(10^{n+1}-9n-10).

The kthk^{th} term of the series is 5,55,555,…5, 55, 555, \ldots which can be written as 5×(1,11,111,…)5\times(1,11,111,\ldots).

Each repunit can be written as 10k−19\dfrac{10^k-1}{9} (e.g. 102−19=999=11\dfrac{10^2-1}{9}=\dfrac{99}{9}=11).

So the kthk^{th} term =5×10k−19=5(10k−1)9= 5\times\dfrac{10^k-1}{9} = \dfrac{5(10^k-1)}{9}.

Sn=∑k=1n5(10k−1)9=59[∑k=1n10k−n]S_n = \sum_{k=1}^{n}\dfrac{5(10^k-1)}{9} = \dfrac{5}{9}\left[\sum_{k=1}^{n}10^k - n\right]

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

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