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If `a_1,a_2,a_3, ,a_n`are an A.P. ofnon-zero terms, prove that`1/(a_1+a_2)+1/(a_1+a_3)++1/(a_(n-1)+a_n)=(n-1)/(a_1+a_n)dot` |
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Answer» Let a be the common difference of the given AP. Then `(a_(2) a_(1)) = (a_(3) -a_(2)) = (a_(4) - a_(3)) = …..= ( a_(n) -a_(n-1)) =d` ` 1/ (a_(1)a_(2)) + 1/(a_(2) a_(3)) + 1/(a_(3) a_(4)) +….+ 1/(a_(n-1) a_(n)) ` ` = 1/d . { d/(a_(1)a_(2)) + d/(a_(2)a_(3)) + d/ (a_(3) a_(4)) +.....+ d/(a_(n-1) a_(n)) }` ` =1/d .{ ((a_(2) -a_(1))/(a_(1)a_(2)) + ((a_(3) -a_(2))/(a_(2) a_(3)) + ((a_(4) -a_(3)))/(a_(3)a_(4)) +....+((a_(n) -a_(n-1)))/(a_(n-1)a_(n))}` `[ because(a_(2)-a_(1))=(a_(3)-a_(2))=....=(a_(n)-a_(n-1))=d]` `=1/d.{(1/a_(1)-1/a_(2))+(1/a_(2)-1/a_(3))+(1/a_(3)-1/a_(4))+....+(1/a_(n-1) -1/a_(n))}` `= 1/d .[ ({a_(1) + (n-1) d} -a_(1))/(a_(1)a_(n))]` ` 1/d. ((n-1)d)/(a_(1)a_(n))= ((n-1))/(a_(1) a_(n))` Hence, the result follows. |
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