Question

if 1st 2nd and last term of an A.P. is a, b and c respectively prove that the sum is (a+c) (b+c-2a) / 2(b-a)

Answer 1

here
first term=a,
common difference (d)=b-a
Tn=c,

a+(n-1)d=c,

then

(n-1)d=c-a,

(n-1)=(c-a)/d,

then

n= (c-a)/d + 1,

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

n=[c-a+b-a]/(b-a),

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


therefore

Sum=(b+c-2a)/[2(b-a)] × [2a+c-a],

Sum =(b+c-2a)/2(b-a) ×(a+c)