The sum of three consecutive terms in an arithmetic sequence is 30, and their product is 750. Find the three terms. (Suggestion: Let x denote the middle term and d the common difference.)
A) 5, 9, 13
B) 4, 9, 14
C) 5, 10, 15
D) 1, 10, 15
E) 6, 10, 14

Question

Quizplus · Accepted Answer

Let's use the suggested variables. 
We know that the sum of three consecutive terms is 30, therefore:

x-1 + x + x+1 = 30 
3x = 30 
x = 10

So the middle term is 10.

To find the first and third terms, we can use the fact that their product is 750:

(x-1)*x*(x+1) = 750

9*10*11 = 990, which is greater than 750, so we know that the smallest term must be smaller than 9.

7*8*9 = 504, which is too small.

Therefore, the smallest term must be 5:

4*5*6 = 120, which is too small.

5*6*7 = 210, which is also too small.

5*10*15 = 750, which is the desired product.

So the three terms are 5, 10, and 15, which is choice C.

The Sum of Three Consecutive Terms in an Arithmetic Sequence