InterviewSolution
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InterviewSolution
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If `a_1, a_2, a_3, ,a_(2n+1)`are in A.P., then`(a_(2n+1)-a_1)/(a_(2n+1)+a_1)+(a_(2n)-a_2)/(a_(2n)+a_2)++(a_(n+2)-a_n)/(a_(n+2)+a_n)`is equal to`(n(n+1))/2xx(a_2-a_1)/(a_(n+1))`b. `(n(n+1))/2`c. `(n+1)(a_2-a_1)`d. none of theseA. `(n(n+1))/(2)*(a_(2)-a_(1))/(a_(n+1))`B. `(n(n+1))/(2)`C. `(n+1)(a_(2)-a_(1))`D. none of these |
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Answer» Correct Answer - A In an A.P. the sum of the terms equidistant from the beginning and the end is always same is equal to the sum of first and last term. Therefore, `a_(1)+a_(2n+1)=a_(2)+a_(2n)= . . .=a_(n)+a_(n+2)` `:." "(a_(2n+1)-a_(1))/(a_(2n+1)+a_(1))=(a_(2n)-a_(2))/(a_(2n)+a_(2))+ . . .+(a_(n+2)-a_(n))/(a_(n+2)+a_(n))` `=(2nd+(2n-2)d+ . . .+2d)/(a_(2n+1)+a_(1))` `=2d(n(n+1))/(2)xx(1)/(a_(2n+1)+a_(1))` `=2d(n(n+1))/(2)xx(1)/(2a_(n+1))" "[becausea_(1)+a_(2n+1)=a_(n)+a_(n+2)=2a_(n+1)]` `=d(n(n+1))/(2)xx(1)/(a_(n+1))=(n(n+1))/(2)*(a_(2)-a_(1))/(a_(n+1))" "[becaused=a_(2)-a_(1)]` |
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