# 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?

Jul 17, 2015

The sum of the first 22 terms is $\textcolor{red}{\frac{2453}{2}}$.

#### Explanation:

We know that ${a}_{3} = 9$ and ${a}_{7} = 31$.

We also know that ${a}_{n} = {a}_{1} + \left(n - 1\right) d$.

${a}_{7} = {a}_{1} + \left(7 - 1\right) d$

Equation (1): $31 = {a}_{1} + 6 d$

and

${a}_{3} = {a}_{1} + \left(3 - 1\right) d$

Equation (2): $9 = {a}_{1} + 2 d$

Subtract Equation (2) from Equation (1).

$31 - 9 = 6 d - 2 d$

$22 = 4 d$

$d = \frac{22}{4}$

Equation (3): $d = \frac{11}{2}$

Substitute Equation (3) in Equation (1).

$31 = {a}_{1} + 6 d$

31 = a_1 + 6×11/2 = a_1 + 33

${a}_{1} = 31 - 33 = - 2$

So ${a}_{1} = - 2$ and $d = \frac{11}{2}$.

${a}_{n} = {a}_{1} + \left(n - 1\right) d$

So the 22nd term is given by

a_22 = -2 + (22-1)×11/2 = -2 + 21×11/2 = -4/2 + 231/2

${a}_{22} = \frac{227}{2}$

The sum ${S}_{n}$ of the first $n$ terms of an arithmetic series is given by

${S}_{n} = \frac{n \left({a}_{1} + {a}_{n}\right)}{2}$

So

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

${S}_{22} = \frac{2453}{2}$