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:

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

Solution

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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1955 AHSME Problems/Problem 45

Problem 45

Solution