2007 AMC 12A Problem 7

Attempt Problem 7 of the 2007 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Let

a,

b,

c,

d,

and

e

be five consecutive terms in an arithmetic sequence, and suppose that

a+b+c+d+e=30.

Which of the following can be found?

a

b

c

d

e

Write the five terms as

c-2D,\ c-D,\ c,

\ c+D,\ c+2D

for a common difference

D

The sum collapses to

5c,

while the outer terms still depend on the unknown

D

Let

D

be the common difference. Then

a=c-2D,

b=c-D,

d=c+D,

and

e=c+2D,

a+b+c+d+e=5c.

Thus

5c=30,

giving

c=6.

The other terms cannot be determined: the sequences

4,5,6,7,8

and

10,8,6,4,2

both satisfy the conditions but differ in every term except the middle one.

Thus, the correct answer is C.

Problem 7 in other years