If a1,a2,a3,…,an are in arithmetic progression, where a1>0 for all i.
Prove that 1√a1+√a2+1√a2+√a3+…+1√an−1+√an=n−1√a1+√an
Since a1,a2,a3,…,an are in AP.
∴ (a2−a1)=(a3−a2)=…=(an−an−1) =d(say)
[∵ common difference of an AP are equal]
Now, LHS=1√a1+√a2+1√a2+√a3+…+1√an−1+√an
=√a2−√a1a2−a1+√a3−√a2a3−a2+…+√an−√an−1an−an−1
[rationalise each term]
=√a2−√a1d+√a3−√a2d+…+√an−√an−1d
=1d[√a2−√a1+√a3−√a2+…+√an−√an−1]
=1d[√an−√a1]=an−a1d[√an+√a1]
[∵ multiplying numerator and denominator by √an+√a1]
=[a1+(n−1)d−a1]d(√an+√a1) [∵an=a1+(n−1)d]
=(n−1)dd(√an+√a1)=n−1√a1+√an=RHS