1974 AMC 12 Problem 29

Question

For

p=1,

2,

\ldots,

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

80{,}000

80{,}200

80{,}400

80{,}600

80{,}800

Use the

40

-term arithmetic-series formula for a fixed

p

Simplify

S_p=20\bigl(2p+39(2p-1)\bigr)

before summing over

p

For each

p,

\begin{aligned} S_p&=\frac{40}{2} \bigl(2p+39(2p-1)\bigr)\\ &=1600p-780. \end{aligned}

Since

1+2+\cdots+10=55,

therefore

\begin{aligned} \sum_{p=1}^{10}S_p &=1600(55)-7800\\ &=80{,}200. \end{aligned}

Therefore, the correct answer is B.

Problem 29 in other years