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?

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Answer 1 · 2015-07-17T21:41:37

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

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#