1974 AHSME Problems/Problem 29

Problem

For $p=1, 2, \cdots, 10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1$ ; then $S_1+S_2+\cdots+S_{10}$ is

$\mathrm{(A)\ } 80000 \qquad \mathrm{(B) \ }80200 \qquad \mathrm{(C) \ } 80400 \qquad \mathrm{(D) \ } 80600 \qquad \mathrm{(E) \ }80800$

Solution

The $40\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$ . Therefore, the sum of the first $40$ terms of such a progression is $\frac{40}{2}(79p-39+p)=1600p-780$ .

We now want to evaluate $\sum_{p=1}^{10}(1600p-780)$ . $\sum_{p=1}^{10}(1600p-780)=1600\sum_{p=1}^{10}(p)-\sum_{p=1}^{10}(780)$ $=(1600)\left(\frac{10\cdot11}{2}\right)-(780)(10)=88000-7800=80200, \boxed{\text{B}}.$

1974 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1974 AHSME Problems/Problem 29

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