The first , second and the last terms of an A.P. are `a ,b , c` respectively. Prove that the sum is `((a+c)(b+c-2a))/(2(b-a))` .

Explanation

`1^(st)` term=a

`2^(nd)`term=b

`3^(rd)`term=c

Common diff,d=(b-a)

c=a+(n-1)(b-a)

c-a=(n-1)(b-a)

`n-1=(c-a)/(b-a)`

`n=(c-a)/(b-a)+1`

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