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

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

$S_22 = 2453/2$