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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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 $\frac{2453}{2}$ $color(red)(2453/2)$ .

Explanation:

We know that $a_{3} = 9$ $a_3 = 9$ and $a_{7} = 31$ $a_7 = 31$ .

We also know that $a_{n} = a_{1} + (n - 1) d$ $a_n = a_1 + (n-1)d$ .

$a_{7} = a_{1} + (7 - 1) d$ $a_7 = a_1 + (7-1)d$

Equation (1): $31 = a_{1} + 6 d$ $31 = a_1 + 6d$

and

$a_{3} = a_{1} + (3 - 1) d$ $a_3 = a_1 + (3-1)d$

Equation (2): $9 = a_{1} + 2 d$ $9 = a_1 + 2d$

Subtract Equation (2) from Equation (1).

$31 - 9 = 6 d - 2 d$ $31-9 = 6d -2d$

$22 = 4 d$ $22 = 4d$

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

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

Substitute Equation (3) in Equation (1).

$31 = a_{1} + 6 d$ $31 = a_1 + 6d$

$31 = a_{1} + 6 \times \frac{11}{2} = a_{1} + 33$ $31 = a_1 + 6×11/2 = a_1 + 33$

$a_{1} = 31 - 33 = - 2$ $a_1 = 31-33 = -2$

So $a_{1} = - 2$ $a_1 = -2$ and $d = \frac{11}{2}$ $d = 11/2$ .

$a_{n} = a_{1} + (n - 1) d$ $a_n = a_1 + (n-1)d$

So the 22nd term is given by

$a_{22} = - 2 + (22 - 1) \times \frac{11}{2} = - 2 + 21 \times \frac{11}{2} = - \frac{4}{2} + \frac{231}{2}$ $a_22 = -2 + (22-1)×11/2 = -2 + 21×11/2 = -4/2 + 231/2$

$a_{22} = \frac{227}{2}$ $a_22 = 227/2$

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

$S_{n} = \frac{n (a_{1} + a_{n})}{2}$ $S_n = (n(a_1+a_n))/2$

So

$S_{22} = \frac{22 (a_{1} + a_{22})}{2} = \frac{22}{2} \times (- 2 + \frac{227}{2}) = 11 \times (- \frac{4}{2} + \frac{227}{2}) = 11 \times \frac{223}{2}$ $S_22 = (22(a_1+a_22))/2 = 22/2×(-2+227/2) = 11×(-4/2 + 227/2) = 11 × 223/2$

$S_{22} = \frac{2453}{2}$ $S_22 = 2453/2$

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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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