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!