1955 AHSME Problems/Problem 45

Problem 45

Given a geometric sequence with the first term $\neq 0$ and $r \neq 0$ and an arithmetic sequence with the first term $=0$. A third sequence $1,1,2\ldots$ is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:

$\textbf{(A)}\ 978\qquad\textbf{(B)}\ 557\qquad\textbf{(C)}\ 467\qquad\textbf{(D)}\ 1068\\ \textbf{(E)}\ \text{not possible to determine from the information given}$


Let our geometric sequence be $a,ar,ar^2$ and let our arithmetic sequence be $0,d,2d$. We know that \[\begin{cases} a+0=1 \\ ar+d=1\\ ar^2+2d=2\end{cases}\] This implies that $a=1$, hence $r+d=1$ and $r^2+2d=2$. We can rewrite $r+d=1$ as $d = 1-r$, plugging this into $r^2+2d=2$, we get $r^2+2-2r = 2$, we can simplify this to get $r(r-2)=0$, so $r=0$ or $2$. But since $r\neq 0$, $r=2$ and $d=-1$. So our two sequences are $\{1-n\}$ and $\{2^{n-1}\}$, which means the third sequence will be \[\{2^{n-1}+n+1\}=\{1,1,2,5,12,27,\dots\}\]Calculating the sum of the first 10 terms and adding them up yields 978, hence our answer is $\fbox{A}$.

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