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Q;if a1,a2,a3....an are positive numbers in a.p. then show that 1/root(a1)+root(a2) +1/root(a2)+root(a3) + 1/root(a3)+root(a4).....+ 1/root(an)+root(an+1)= n/root(a1)+root(an+1)
Expert Reply
let d be the common difference in the given AP
So if we take the first term
1/{root(a1) + root(a2)} and rationalize the denominator taht is multiply the numerator and the denominator by root(a2)  root(a1)
we get ,
{root(a2)  root(a1)}/(a2a1) = {root(a2)  root(a1)}/d
Similarly 1/{root(a2) + root(a3)} = {root(a3)  root(a2)}/d
On adding all the terms we get
{root(a2)  root(a1) + root(a3)  root(a2) + ......... + root(an+1)  root(an)}/d
= {root(an+1)  root(a1)}/d .........1
Also d = {a_{n+1  }a1}/n = {(root(an+1)  root(a1))(root(an+1) + root(a1))}/n
On substituting the value of d in eq1 we get
R.H.S.
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