Question

in first term of A.P is a,second term is b and last

Answer 1

first term of AP=a 
then second term must be a+d
it is given as b
then a+d =b
       common difference d=b-a
last term=a+(n-1)d
  it is given as c
c=a+(n-1)(b-a)
 c-a=(n-1)[b-a]
[c-a/b-a]+1=n
n=c+b-2a/b-a
sum of all terms=n/2[a+l]
                          =(c+b-2a/b-a)/2([a+c])
                            =(c+b-2a)(a+c)/2(b-a)