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Q.Find the sum of the series 0.7+0.77+0.777+…0.7 + 0.77 + 0.777 + \ldots to nn terms. OR The sum of two numbers is 6 times their geometric mean, show that the numbers are in the ratio (3+22):(3−22)(3+2\sqrt2):(3-2\sqrt2).

Nagaland NbseNagaland Board of School Education (Class XI) 2023Subjective· 6mImportance★★★★★
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Primary: factor out 7/9 and sum the resulting geometric series. OR alternative: turn "sum = 6×GM" into a quadratic in the ratio, then match to the target form by rationalizing.

Primary — sum of the series:

Sn=0.7+0.77+0.777+⋯S_n=0.7+0.77+0.777+\cdots to nn terms =7[0.1+0.11+0.111+⋯ ]=79[0.9+0.99+0.999+⋯ ]=7[0.1+0.11+0.111+\cdots]=\dfrac79[0.9+0.99+0.999+\cdots]

=79[(1−0.1)+(1−0.01)+⋯+(1−10−n)]=79[n−(0.1+0.01+⋯+10−n)]=\dfrac79\left[(1-0.1)+(1-0.01)+\cdots+(1-10^{-n})\right]=\dfrac79\left[n-\left(0.1+0.01+\cdots+10^{-n}\right)\right].

The bracketed geometric sum =0.1(1−0.1n)1−0.1=1−10−n9=\dfrac{0.1(1-0.1^n)}{1-0.1}=\dfrac{1-10^{-n}}{9}.

So Sn=79[n−1−10−n9]=7n9−781(1−10−n)S_n=\dfrac79\left[n-\dfrac{1-10^{-n}}{9}\right]=\dfrac{7n}{9}-\dfrac{7}{81}(1-10^{-n}).

OR alternative — GM ratio:

Let the numbers be a,ba,b with ratio x=abx=\dfrac ab. Given a+b=6aba+b=6\sqrt{ab}, divide by bb: x+1=6xx+1=6\sqrt x.

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