finding sum of first 22 terms in an arithmetic sequence when 3rd term is 9 and 7th term is 31

Question

third term of sequence is 9

seventh term of sequence is 31

what is the sum of the first 22 terms?

Answer 1

In arithmetic sequence sum =n/2(2a+(n-1)d)

Given third term is a+2d = 9

seventh term is a+6d = 31

Subtract the 7th term from 3rd term.

a + 2d   = 9

a + 6d  = 31

(-) (-)    (-)

_________

-4d = -22

Divide by 4 to each side.and cancell the signs.

4d/4 = 22/4

d = 5.5

Substitute the d value in 3rd term.

a+2*5.5 = 9

a+11 = 9

Subtract 11from each side.

a+11-11= 9-11

a =9-11

a =-2

and n =22

Now sum of 22 terms = 22/2[(2*-2)+(22-1)5.5]

= 11(-4+21*5.5)

= 11(-4+115.5)

= 11*111.5

= 1226.5