Sn=1/2[ 7n2+9n]?

Question

the sum of first n terms in an arithmetic progressions( the above one) find the first term and the common difference

Answer 1

The n terms of the arithmetical progression are
A, A+D, A+2D, A+3D, ..., A + (n-1)D
So the last term is L = A + (n-1)D
The sum is Sn = (1/2)(n)(A + L), which is given to be (1/2)(7n2+9n).
So we get
(1/2)(n)(A + L) = (1/2)(7n2+9n) = (1/2)(n)(7n+9)
A + L = 7n + 9
A + A + (n-1)D = 7n + 9
2A + Dn - D = 7n + 9
Dn + (2A - D) = 7n + 9
From this we get
D = 7 and
2A - D = 9
So 2A - 7 = 9, and A = 8

So the first term is A = 8, the difference is D = 7, and the last term is L = 8 + 7(n-1).
To check the answer,
Sn = (1/2)(n)(A + L) = (1/2)(n)(8 + 8 + 7(n-1)) = (1/2)(n)(7n + 9)

Answer 2

Sn = n/2* [2*a1 + (n-1)*d];

Sn = 1/2* [2*a1*n + (n^2-n)*d]

Sn = 1/2* [(2*a1*-d)n + d*n^2]

by comparison with your equation, d = 7 and 2*a1-7 = 9, 2*a1 = 16
a1 = 8

First term is 8, common difference = 7

Answer 3

Hold it, what do you do in the maths class? Give it a try yourself so that you can learn something!