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If a1,a2,a3, ,an are in A.P., where ai >...

If `a_1,a_2,a_3, ,a_n` are in A.P., where `a_i >0` for all `i` , show that `1/(sqrt(a_1)+sqrt(a_2))+1/(sqrt(a_1)+sqrt(a_3))++1/(sqrt(a_(n-1))+sqrt(a_n))=(n-1)/(sqrt(a_1)+sqrt(a_n))dot`

Text Solution

Verified by Experts

L.H.S. `= (1)/(sqrt(a_(1)) + sqrt(a_(2))) + (1)/(sqrt(a_(2)) + sqrt(a_(3))) + ....+ (1)/(sqrt(a_(n - 1)) + sqrt(a_(n)))`
`= (1)/(sqrt(a_(2)) + sqrt(a_(1))) + (1)/(sqrt(a_(3)) + sqrt(a_(2))) + .... + (1)/(sqrt(a_(n)) + sqrt(a_(n - 1)))`
`=(sqrt(a_(2)) - sqrt(a_(1)))/((a_(2) - a_(1))) + (sqrt(a_(3)) - sqrt(a_(2)))/((a_(3) - a_(2))) + ... + (sqrt(a_(n))- sqrt(a_(n - 1)))/(a_(n ) - a_(n - 1))`
Let 'd' is the common difference of this A.P.
then `a_(2) - a_(1) = a_(3) - a_(2) = ....... = a_(n) - a_(n -1) = d`
Now L.H.S.
`= (1)/(d) {sqrt(a_(2)) - sqrt(a_(1)) + sqrt(a_(3)) - sqrt(a_(2)) + .... + sqrt(a_(n - 1)) - sqrt(a_(n - 2)) + sqrt(a_(n - 1))} = (1)/(d) {sqrt(a_(n)) - sqrt(a_(1))}`
`= (a_(n) - a_(n))/(d(sqrt(a_(n)) + sqrt(a_(1)))) = (a_(1) + (n - 1) d- a_(1))/(d(sqrt(a_(n)) + sqrt(a_(1)))) = (1)/(d) ((n - 1)d)/(d(sqrt(a_(n)) + sqrt(a_(1)))) = (n - 1)/(sqrt(a_(n)) + sqrt(a_(1))) =` R.H.S
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