Question

Show that the sum of a A.P. whose first term is 'a', the second term is 'b' and the last term is 'c', is equal to (a+c)(b+c-2a)/2(b-a)

Answer 1

First term =a
second term =b
last term =c
common difference=(b-a)
tn =a+(n-1) d

c=a+(n-1) d

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

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

( c+b-2a)/(b-a)=n--------------------(1)

now ,
we know sum of n terms=n/2 (first term+last term)

put equation (1)value
= (b+c-2a)/(b-a)(a+c)

hence Sn=[(b+c-2a)(c+a)/(b-a)]