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 24532.

Explanation:

We know that a3=9 and a7=31.

We also know that an=a1+(n1)d.

a7=a1+(71)d

Equation (1): 31=a1+6d

and

a3=a1+(31)d

Equation (2): 9=a1+2d

Subtract Equation (2) from Equation (1).

319=6d2d

22=4d

d=224

Equation (3): d=112

Substitute Equation (3) in Equation (1).

31=a1+6d

31=a1+6×112=a1+33

a1=3133=2

So a1=2 and d=112.

an=a1+(n1)d

So the 22nd term is given by

a22=2+(221)×112=2+21×112=42+2312

a22=2272

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

Sn=n(a1+an)2

So

S22=22(a1+a22)2=222×(2+2272)=11×(42+2272)=11×2232

S22=24532