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Question
If a1, a2, a3, ..., an are in A.P., where ai > 0 for all i, show that `1/(sqrt(a_1) + sqrt(a_2)) + 1/(sqrt(a_2) + sqrt(a_3)) + ... + 1/(sqrt(a_(n - 1)) + sqrt(a_n)) = (n - 1)/(sqrt(a_1) + sqrt(a_n))`
Solution
Given that a1, a2, a3, ..., an are in A.P.
∴ Common difference d = a2 – a1
= a3 – a2
= a4 – a3
= …
= an – an–1
If a2 – a1 = d then `sqrt(a_2^2) - sqrt(a_1^2)` = d
⇒ `(sqrt(a_1) - sqrt(a_1)) (sqrt(a_2) + sqrt(a_1))` = d .....[∵ a2 – b2 = (a + b)(a – b)]
⇒ `1/(sqrt(a_1) + sqrt(a_2)) = (sqrt(a_2) - sqrt(a_1))/d`
Similarly `1/(sqrt(a_2) + sqrt(a_3)) = (sqrt(a_3) - sqrt(a_2))/d`
`1/(sqrt(a_3) + sqrt(a_4)) = (sqrt(a_4) - sqrt(a_3))/d`
.... .... ....
`1/(sqrt(a_(n - 1)) + sqrt(a_n)) = (sqrt(a_n) - sqrt(a_(n - 1)))/d`
Adding the above terms, we get
`1/(sqrt(a_1) + sqrt(a_2)) + 1/(sqrt(a_2) + sqrt(a_3)) + 1/(sqrt(a_3) + sqrt(a_4)) + ... + 1/(sqrt(a_(n - 1)) + sqrt(a_n))`
= `1/d[sqrt(a_2) - sqrt(a_1) + sqrt(a_3) - sqrt(a_4) - sqrt(a_3) + ... sqrt(a_n) - sqrt(a_(n - 1))]`
= `1/d[sqrt(a_n) - sqrt(a_1)]` ....(i)
Now an = a1 + (n – 1)d
⇒ an – a1 = (n – 1)d
⇒ `sqrt(a_n^2) - sqrt(a_1^2)` = (n – 1)d
⇒ `(sqrt(a_n) + sqrt(a_1))(sqrt(a_n) - sqrt(a_1))` = (n – 1)d
⇒ `sqrt(a_n) - sqrt(a_1) = ((n - 1)d)/(sqrt(a_n) + sqrt(a_1))`
⇒ `(sqrt(a_n) - sqrt(a_1))/d = (n - 1)/(sqrt(a_n) + sqrt(a_1))` ....(ii)
From equation (i) and equation (ii) we get
`1/(sqrt(a_1) + sqrt(a_2)) + 1/(sqrt(a_2) + sqrt(a_3)) + 1/(sqrt(a_3) + sqrt(a_4)) + ... + 1/(sqrt(a_(n - 1)) + sqrt(a_n)) = (n - 1)/(sqrt(a_n) + sqrt(a_1))`
Hence proved.
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