The arithmetic sequence $a, a+d$, $a+2 d, a+3 d, \ldots, a+(n-1) d$ has the following properties:

- When the first, third, and fifth, and so on terms are added, up to and including the last term, the sum is 320 .
- When the first, fourth, seventh, and so on, terms are added, up to and including the last term, the sum is 224 .

What is the sum of the whole sequence?
(A) 656
(B) 640
(C) 608
(D) 704
(E) 672
When we start at the first term and look at every other term from there, we will look at the last term. Therefore, the total number of terms after the first is a multiple of 2.

When we start at the first term and look at every third term from there, we will look at the last term. Therefore, the total number of terms after the first is a multiple of 3.

Thus, the total number of terms after the first must be a multiple of 6 , so the total number of terms in the sequence can be written as $6 k+1$.

Now we look at the given sums, knowing that $n=6 k+1$.

When we add up every other term, including the first and the last, we are adding up $3 k+1$ terms in total.

Therefore, $\frac{1}{2}(3 k+1)(a+a+6 k d)=320$ or $(3 k+1)(2 a+6 k d)=640$.

(The sum of an arithmetic sequence is half of the product of the number of terms and the sum of the first and last terms. The sequence we get by looking at every other term (or every third term) of an arithmetic sequence is again arithmetic.)

When we add up every third term, including the first and the last, we are adding up $2 k+1$ terms in total.

Therefore, $\frac{1}{2}(2 k+1)(a+a+6 k d)=224$, or $(2 k+1)(2 a+6 k d)=448$.

Dividing the first equation by the second, we obtain $\frac{3 k+1}{2 k+1}=\frac{640}{448}=\frac{10}{7}$, so $k=3$.

Thus, $(3(3)+1)(2 a+6 k d)=640$ or $2 a+6 k d=64$.

We want to determine the sum of the entire arithmetic sequence, which is

$$
\frac{1}{2}(6 k+1)(a+a+6 k d)=\frac{1}{2}(19)(2 a+6 k d)=\frac{1}{2}(19)(64)=608
$$

ANSWER: (C)