# 1973 AHSME Problems/Problem 26

## Problem

The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is $\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$

## Solution

Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference. That means the last term is $a+r(n-1)$.

We can write an equation on the difference between the last and first term based on the conditions. $$a+r(n-1)-a =10.5$$ $$rn-r=10.5$$ Also, half of the terms add up to $24$ while the other half of the terms add up to $30$, so $$24 + r\frac{n}{2} = 30$$ $$nr = 12$$ Substituting the value back to a previous equation, $$12-r=10.5$$ $$r=1.5$$ Substituting to a previous equation again, $$1.5n-1.5=10.5$$ $$n=8$$ Thus, there are $\boxed{\textbf{(E)}\ 8}$ terms in the arithmetic sequence.