The third term of an arithmetic sequence is 9 and the seventh term is 31, how do you find the sum of the first twenty-two terms?

1 Answer
Jul 17, 2015

The sum of the first 22 terms is color(red)(2453/2).

Explanation:

We know that a_3 = 9 and a_7 = 31.

We also know that a_n = a_1 + (n-1)d.

a_7 = a_1 + (7-1)d

Equation (1): 31 = a_1 + 6d

and

a_3 = a_1 + (3-1)d

Equation (2): 9 = a_1 + 2d

Subtract Equation (2) from Equation (1).

31-9 = 6d -2d

22 = 4d

d = 22/4

Equation (3): d= 11/2

Substitute Equation (3) in Equation (1).

31 = a_1 + 6d

31 = a_1 + 6×11/2 = a_1 + 33

a_1 = 31-33 = -2

So a_1 = -2 and d = 11/2.

a_n = a_1 + (n-1)d

So the 22nd term is given by

a_22 = -2 + (22-1)×11/2 = -2 + 21×11/2 = -4/2 + 231/2

a_22 = 227/2

The sum S_n of the first n terms of an arithmetic series is given by

S_n = (n(a_1+a_n))/2

So

S_22 = (22(a_1+a_22))/2 = 22/2×(-2+227/2) = 11×(-4/2 + 227/2) = 11 × 223/2

S_22 = 2453/2