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Q.Find the sum to nn terms of the series 5+55+555+…5+55+555+\dots OR If A and G be A.M and G.M respectively between two positive numbers, prove that the numbers are A±(A+G)(A−G)A \pm \sqrt{(A+G)(A-G)}.

Nagaland NbseNagaland Board of School Education (Class XI) 2022Subjective· 6mImportance★★★★★
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(a) Rewrite each term of 5+55+555+…5+55+555+\dots as 5×5\times a repunit and sum the resulting geometric series. Or (b) form the quadratic whose roots are the two numbers (using their AM and GM) and solve it.

(a) Sum to nn terms of 5+55+555+…5+55+555+\dots

The kthk^{\text{th}} term is 5,55,555,⋯=5×(1,11,111,… )=5×10k−195,55,555,\dots = 5\times(1,11,111,\dots) = 5\times\dfrac{10^k-1}{9}

Sn=∑k=1n5⋅10k−19=59(∑k=1n10k−n)S_n=\displaystyle\sum_{k=1}^n 5\cdot\dfrac{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\displaystyle\sum_{k=1}^n 10^k = \dfrac{10(10^n-1)}{10-1}=\dfrac{10(10^n-1)}{9}

Sn=59(10(10n−1)9−n)=50(10n−1)81−5n9S_n=\dfrac{5}{9}\left(\dfrac{10(10^n-1)}{9}-n\right)=\dfrac{50(10^n-1)}{81}-\dfrac{5n}{9}

(Check n=1n=1: 50(9)81−59=45081−59=509−59=5\dfrac{50(9)}{81}-\dfrac{5}{9}=\dfrac{450}{81}-\dfrac{5}{9}=\dfrac{50}{9}-\dfrac{5}{9}=5 ✓. Check n=2n=2: gives 60=5+5560=5+55 ✓.)

Or (b) If A, G are the A.M and G.M of two positive numbers, prove the numbers are A±(A+G)(A−G)A\pm\sqrt{(A+G)(A-G)}

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