InterviewSolution
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InterviewSolution
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If `a_r>0, r in N` and `a_1.a_2,....a_(2n)` are in A.P then `(a_1+a_2)/(sqrta_1+sqrta_2)+(a_2+a_(2n-1))/(sqrta_2+sqrta_3)+.....+(a_n+a_(n+1))/(sqrt a_n+sqrta_(n+1))=`A. n-1B. `(n(a_(1)+a_(2n)))/(sqrt(a_(1))+sqrt(a_(n+1)))`C. `(n-1)/(sqrt(a_(1))+sqrt(a_(n+1)))`D. none of these |
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Answer» Correct Answer - B We have, `a_(1)+a_(2n)=a_(2)+a_(2n-1)=a_(3)+a_(2n-2)= . . . =a_(n)+a_(n+1)` Let d be the common difference of the given A.P. Then, Given expression `=(a_(1)+a_(2n)){(1)/(sqrt(a_(1))+sqrt(a_(2)))+(1)/(sqrt(a_(2))+sqrt(a_(3)))+ . . .+(1)/(sqrt(a_(n))+sqrt(a_(n+1)))}` `=((a_(1)+a_(2n)))/(-d){sqrt(a_(1))-sqrt(a_(2))+sqrt(a_(2))-sqrt(a_(3))+ . . .+(1)/(sqrt(a_(n))-sqrt(a_(n+1)))}` `=((a_(1)+a_(2n)))/(d){sqrt(a_(n))-sqrt(a_(n+1))}` `=-((a_(1)+a_(2n)))/(d)xx(a_(1)-a_(n+1))/(sqrt(a_(1))+sqrt(a_(n+1)))` `=(n(a_(1)+a_(2n)))/(sqrt(a_(1))+sqrt(a_(n+1)))" "[becausea_(1)-a_(n+1)=-nd]` |
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