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?

1 Answer
Aug 25, 2016

s_15 = 975

Explanation:

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!