Maths

The 3rd of A.P. sequence=1, 6th is-11,the sum of first 32 term
let first term a, difference d
a-2d=1...(1)
a-5d=-11...(2)
(1)-(2)
3d=12
d=4
a=9
The sum of first 32 term
=(n/2)[2a+(n-1)d]
=(32/2)[2*9+(32-1)*4]
=2272
24. the sum to nth term of an A.P. is n^2-3n, general term=?
The sum to nth term of an A.P. is n^2-3n
S(1)=a=1-3=-2
S(2)=4-6=-2
d=S(2)-2S(1)=2
general term=a+(n-1)d=-2+2(n-1)=2n-4
OR
(n/2)[2a+(n-1)d]=n^2-3n
[2a+(n-1)d]=2(n-3)
a+a+(n-1)d=2(n-3)
-2+a+(n-1)d=2(n-3)
a+(n-1)d=2(n-3)+2=2n-4
general term=2n-4