What is the sum of the described series? Use . Sn=(a1+an/2)

A company has agreed to make a yearly monetary gift to a charity. The first year it donates $2,000. Each year after, it promises to increase the donation by $500. How much total money will the company have given to the charity after 15 years?

A.
$82,500
B.
$77,000
C.
$9,000
D.
$5,500

The sequence of gifts is an arithmetic sequence with a first term of $2000 and a common difference of $500. These values can be used to find the amount of the n-th gift. With that amount, we are able to use the given formula to find the sum of the gifts.

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n-th term

The general term of an arithmetic sequence is ...

an = a1 +d(n -1) . . . . . where a1 is the first term, and d is the common difference

Our sequence has a1=2000, and d=500, so the n-th term is ...

an = 2000 +500(n -1)

The 15th term is ...

a15 = 2000 +500(15 -1) = 9000

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Sum

The problem statement tells us the sum of n terms of an arithmetic series is ...

Sn = (a1 +an)n/2

We have a 15-term series with a1=2000, an=9000, so the sum is ...

S15 = (2000 +9000)(15/2) = 165000/2 = 82,500

The company will have given a total of $82,500 to charity after 15 years.

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Additional comment

You can make an appropriate guess at the answer by considering the "parity" of the sum of terms. The first term is 2000, an even thousand. The second term will be 2500, ending in 500, and the third term will be 3000, again an even thousand.

That is, the second term of each additional pair will be an even thousand, but the sum of the two terms of a pair will be a total that ends in 500. If the number of additional pairs after the first term is odd, then the total given to charity will end in 500. Our series has 15 terms, which is 7 pairs in addition to the first term. 7 is odd, so we know the total amount given will end in 500. Only one answer choice matches that description: 82,500.

(5500 is the sum of terms 2 and 3, so we can ignore that choice.)

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Another way to consider this is to look at the sums of pairs of terms at the ends of the sequence. The first and last terms will both end in 000, so their sum will be an even thousand. The 2nd and next-to-last terms will both end in 500, so their sum will be an even thousand. (These sums are each 11,000.) We find that there are 7 such pairs and one term at the center of the sequence that ends in 500, so the total must be a number greater than 77000 that ends in 500.