 # a1,a2,a3,-------an are in A.P thena1a2+a2a3+--------an-1an=

11 years ago

Dear  Pulipaka,

If d is the common difference for the AP, then:
1 / a[k] - 1 / a[k + 1]
= ( a[k + 1] - a[k] ) / ( a[k] a[k + 1] )
= d / ( a[k] a[k + 1] )

Therefore:
1 / ( a[k] a[k + 1] ) = (1 / d)( 1 / a[k] - 1 / a[k + 1] )

Let s = sum(k = 1, n - 1){ 1 / ( a[k] a[k + 1] ) }.

Then s = (1 / d) sum(k = 1 to n - 1)( 1 / a[k] - 1 / a[k + 1] )
= (1 / d) { sum(k = 1 to n - 1)( 1 / a[k] ) - sum(k = 2 to n)(1 / a[k]) }

All terms from k = 2 to k = n - 1 cancel, and:
s = (1 / d){ 1 / a - 1 / a[n] }
= (1 / d)(a[n] - a) / ( a a[n] ).

As a[n] - a = (n - 1)d:
s = (n - 1) / ( a a[n] ).

Best Of luck

Plz Approve the answer...!!!!

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