Question

An arithmetic series consists of consecutive integers that are multiples are of 4. What is the sum of the first nine terms of this sequence if the first term is 0

Answer 1

144

Explanation:

The series is described by the problem statement as ...

... 0 +4 +8 +12 +... +32

= 4*(0 +1 +2 +... +8) . . . . . the first 9 multiples of 4, starting with 0

Now we know the sum of integers 1..n is given by n(n+1)/2. So for n=8, this is ...

... 8·9/2 = 36

Multiplying this by 4, we find the series sum to be ...

= 4·(sum of integers 0 .. 8) = 4·36 = 144

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Comment on the answer choices

It appears your answer choices may not include the value 144. That is unfortunate. You may want to ask your teacher to show you how to work this problem and reconcile the answer choices with the problem statement.

The 9-term series starting with 4 will have sum ...

... 4·(1 +2 +3 +... +9) = 4·(9·10)/2 = 4·45 = 180

Answer 2

144

Explanation:

Remark

Well you could do this the brute force way. 32 is the nineth term in the series. Sum = 0 + 4 + 8 + 12 + ... 32 = 144. That will get you an answer.

This works unless you are asked for the 44434th term. Then you better use a formula. Even the hundredth term would be trouble.

Step One

Find d

d is given as 4

Step 2

find n. There are 9 terms. The 9 is given.

Step 3

Find the ninth term.

Givens

a1 = 0

n = 9

d = 4

Formula for the 9th term: a9 = a1 + (n - 1)*d

Substitute and Solve

a9 = 0 + (9 - 1)*4

a9 = 8*4 = 32

Step 4

Find the sum of the first 9 terms

Sum = (a1 + a9)*n/2

Sum = (0 + 32)*9/2

Sum = 32 * 9/2

Sum = 32 * 4.5

Sum = 144