If a1- a2- a3-an are in G-P- then find the value of -10k-11a2k-a2k-1 if r is common ratio of G-P-

Common ratio =r a _ n = a _ 1 r ^ n 1 1 a_ k ^ 2 a _ k+1 ^ 2 = 1 a _ 1 r ^ 2k 1 a _ 1 r ^ 2k = 1 a _ 1 r ^ 2k 2 (1 r^ 2 ) ∑ _ k ...

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Byju's Answer

Standard X

Mathematics

Properties of GP

If a1, a2, ...

Question

If $a_{1}, a_{2}, a_{3}, . . . a_{n}$ are in G.P. then find the value of $10 \sum k = 1 \frac{1}{a_{k}^{2} - a_{k + 1}^{2}}$ if r is common ratio of G.P.

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Solution

common ratio $= r$
$a_{n} = a_{1} r^{n - 1} \frac{1}{a_{k}^{2} - a_{k + 1}^{2}} = \frac{1}{a_{1} r^{2 k - 1} - a_{1} r^{2 k}} = \frac{1}{a_{1} r^{2 k - 2} (1 - r^{2})} \sum_{k = 1}^{10} . \frac{1}{a_{k}^{2} - a_{k + 1}^{2}} = \frac{1}{a_{1} (1 - r^{2})} [1 + \frac{1}{r^{2}} + 1 \frac{1}{r^{4}} + . . . . + \frac{1}{r^{18}}] = \frac{1}{a_{1} (1 - r^{2})} \frac{(1 - (\frac{1}{r^{2}})^{10})}{(1 - \frac{1}{r^{2}})} = \frac{- (r^{20} - 1)}{a_{1} (1 - r^{2})^{2} . r^{18}} = \frac{(1 - r^{20})}{a_{1} (1 - r^{2})^{2} . r^{18}}$

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If a1,a2,a3,...an are in G.P. then find the value of 10∑k=11a2k−a2k+1 if r is common ratio of G.P.

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