InterviewSolution
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InterviewSolution
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In a sequence of `(4n+1)`terms, the first `(2n+1)`terms are n A.P. whose common difference is 2, and the last `(2n+1)`terms are in G.P. whose commonratio is 0.5 if the middle terms of the A.P. and LG.P. are equal ,then themiddle terms of the sequence is`(n .2 n+1)/(2^(2n)-1)`b. `(n .2 n+1)/(2^n-1)`c. `n .2^n`d. none of theseA. `(n*2^(n+1))/(2^(n)-1)`B. `(n*2^(n+1))/(2^(2n)-1)`C. `n*2^(n)`D. none of these |
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Answer» Correct Answer - A Let `a_(1),a_(2),a_(3), . . . ,a_(2n),a_(2n+1), . . . ,a_(4n),a_(4n+1)` be the given sequence. Then, `a_(2n+1)=a_(1)+4nanda_(4n+a)=a_(2n+1)xx((1)/(2))^(2n)` `rArr" "a_(2n+1)=a_(1)+4nanda_(4n+1)=((a_(1)+2nd))/(2^(2n))` Also, Middle term of first (2n+1) terms = Middle term of the last (2n+1) terms `rArr" "a_(n+1)=a_(3n+1)` `rArr" "a_(1)+2n=a_(2n+1)((1)/(2))^(n)` `rArr" "a_(1)+2n=(a_(1)+4n)xx(1)/(2^(n))` `rArr" "a_(1){1-(1)/(2^(n))}=(4n)/(2^(n))-2n` `rArr" "a_(1)(2^(n)-1)=4n-2nxx2^(n)rArra_(1)=(4n-2nxx2^(n))/(2^(n)-1)` `:.` Middle term of the sequence `=a_(2n+1)=a_(1)+4n` `=(4n-2nxx2^(n))/(2^(n)-1)+4n` `=(2nxx2^(n))/(2^(n)-1)=(n*2^(n+1))/(2^(n)-1)` |
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