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Let `a_1,a_2,a_3….., a_(101)` are in G.P with `a_(101) =25 ` and `Sigma_(i=1)^(201) a_i=625` Then the value of `Sigma_(i=1)^(201) 1/a_i` eaquals _______. |
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Answer» Correct Answer - 1 Let a be the first fterm and r be the common ratio of G.P. Then `a(1-r^(201))/(1-r)=625` (1) Now `sum_(r=1)^(201)1/(a_(i))=1/(a_(1))+1/(a_(2))+..+1/(a_(201))` `=1/a+1/(ar)+..+1/(ar^(200))` `=(1/a((1/r)^(201)-1))/((1/r-1))` `=1/a((1-r^(201))/(1-r))1/r^(200)` `=1/axx625/axx1/r^(200)` [from (1)] `=625/((ar^(100))^(2))` `=625/(a_(101))^(2)` `=625/625=1` |
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