Question

Numeric Response 4. In an arithmetic series, the first term is -12 and the 15th term is 40. The sum of the first 15 terms is (Record your answer in the numerical-response section below.)

Your answer should be in.0000​

Answer 1

In any artihmetic sequence, consecutive terms differ by a fixed constant c. So given the first term a, the second term is a + c, the third terms is a + 2c, and so on, up to the n-th term a + (n - 1)c.

If the 15th term is 40, then

40 = -12 + (15 - 1) c   ==>   c = 52/14 = 26/7

We can then write the n-th term as

-12 + (n - 1) 26/7 = (26n - 110)/7

The sum of the first 15 terms is then

$\displaystyle \sum_{n=1}^{15}\frac{26n-110}7 = \frac{26}7\sum_{n=1}^{15}n - \frac{110}7\sum_{n=1}^{15}n = \boxed{210}$

Another way to compute the sum: let S denote the sum,

S = -12 - 58/7 - 32/7 + … + 228/7 + 254/7 + 40

Reverse the order of terms:

S* = 40 + 254/7 + 228/7 + … - 32/7 - 58/7 - 12

Notice that adding up terms in the same position gives the same result,

-12 + 40 = 28

-58/7 + 254/7 = 28

-32/7 + 228/7 = 28

so that

S + S* = 2S = 28 + 28 + 28 + … + 28 + 28 + 28

There are 15 terms in the sum, so

2S = 15×28   ==>   S = 15×28/2 = 210