How do you find the sum of the first 15 terms of an arithmetic sequence if the first term is 51 and the 25th term is 99?

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Answer 1 · 2016-08-25T15:21:57

$s_15 = 975$

Consider the following example:

The first term of an arithmetic sequence is $2$ and the third is $6$ . What is $d$ , the common difference?

With an arithmetic sequence, the $d$ is added to each term to get the next.

Since $t_1 = 2$ and $t_3 = 6$ , there will be $3 - 1= 2$ $d's$ added to $t_1$ to get $t_3$ . So, we can write the following equation:

$2 + 2d = 6$

$2d = 4$

$d = 2$

It works, too, since if $t_1 = 2$ , $t_2 = 4$ and $t_3 = 6$ , which makes an arithmetic sequence.

The same principle can be applied to our problem.

$25 - 1 = 24$ , so there will be $24 d's$ added to $51$ to get $99$ .

Hence, $51 + 24d = 99$

$24d = 48$

$d = 2$

So, the common difference is $2$ .

All we have to do now is to apply the formula $s_n = n/2(2a + (n - 1)d))$ to determine the sum of the sequence.

$s_15 = 15/2(2(51) + (15 - 1)2)$

$s_15 = 15/2(102 + 28)$

$s_15 = 15/2(130)$

$s_15 = 975$

Thus, the sum of the first fifteen terms in the arithmetic sequence is $975$ .

Hopefully this helps!