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If a(1),a(2),a(3),".....",a(n) are in HP...

If `a_(1),a_(2),a_(3),".....",a_(n)` are in HP, than prove that `a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+"....."+a_(n-1)a_(n)=(n-1)a_(1)a_(n)`

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As `a_(1). A_(2), …, a_(n) ` are in H.P., their reciprocals ` (1)/(a_(1)), (1)/(a_(2)),…, (1)/(a_(n))` are in in A.P. Let d be the common
difference of this A.P.
` (1)/(a_(n)) = (1)/(a_(1)) + (n-1)d `
`rArr (1)/(a_(n)) - (1)/(a_(1)) = (n-1)d`
` rArr a_(1) - a_(n) = (n-1) d. a_(1)a_(n)`
` rArr (1)/(d) (d_(1)_(n)) = (n-1) a_(1) a_(n)` ...(i)
Again ` (1)/(a_(2))- (1)/(a_(1))=(1)/(a_(3)) = (1)/(a_(2)) = ...= (1)/(a_(n) ) - (1)/(a_(n-a)) = d `
`rArr (a_(1) -a_(2))/(a_(1)a_(2) ) = (a_(2) - a_(3))/(a_(2) -a_(3)) = ...= (s_(n-1) -a_(n))/(a_(n-1) -a_(n)) `
` rArr (a_(1) a_(2))/(a_(1)-a_(2))=(a_(2)a_(3))/(a_(2) -a_(3))=...=(a_(n-1) a_(n))/(a_(n-1) -a_(n)) = (1)/(d)`
Now , ` a_(1) a_(2) + a_(2) a_(3) + ...+ a_(n-1) a_(n) = (1)/(d) {(a_(1) -a_(2)) + (a_(2) - a_(3)) + ...+ (a_(n-1) + a_(n))}`
` = (1)/(d) (a_(1) - a_(n))`
` = (n-1)a_(1) a_(n)` [ from (i) ]
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