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)24532.

Explanation:

We know that a_3 = 9a3=9 and a_7 = 31a7=31.

We also know that a_n = a_1 + (n-1)dan=a1+(n1)d.

a_7 = a_1 + (7-1)da7=a1+(71)d

Equation (1): 31 = a_1 + 6d31=a1+6d

and

a_3 = a_1 + (3-1)da3=a1+(31)d

Equation (2): 9 = a_1 + 2d9=a1+2d

Subtract Equation (2) from Equation (1).

31-9 = 6d -2d319=6d2d

22 = 4d22=4d

d = 22/4d=224

Equation (3): d= 11/2d=112

Substitute Equation (3) in Equation (1).

31 = a_1 + 6d31=a1+6d

31 = a_1 + 6×11/2 = a_1 + 3331=a1+6×112=a1+33

a_1 = 31-33 = -2a1=3133=2

So a_1 = -2a1=2 and d = 11/2d=112.

a_n = a_1 + (n-1)dan=a1+(n1)d

So the 22nd term is given by

a_22 = -2 + (22-1)×11/2 = -2 + 21×11/2 = -4/2 + 231/2a22=2+(221)×112=2+21×112=42+2312

a_22 = 227/2a22=2272

The sum S_nSn of the first nn terms of an arithmetic series is given by

S_n = (n(a_1+a_n))/2Sn=n(a1+an)2

So

S_22 = (22(a_1+a_22))/2 = 22/2×(-2+227/2) = 11×(-4/2 + 227/2) = 11 × 223/2S22=22(a1+a22)2=222×(2+2272)=11×(42+2272)=11×2232

S_22 = 2453/2S22=24532