Question

How many terms of the arithmetic sequence 2,4,6,8, ... add up to 60,762?

Answer 1

Let `a_i = 2i` for `i = 1, 2,...`

`60762 = sum_(i=1)^(i=n) a_i = n((a_1+a_n)/2) = n((2+2n)/2) = n(n+1)`

So `n = floor(sqrt(60762)) = 246`

Explanation:

A standard property of sums of arithmetic sequences is:

`sum_(i=1)^(i=n) a_i = n((a_1+a_n)/2)`

Intuitively, the terms of an arithmetic sequence are evenly spaced, so the average value is the same as the average of the two outermost terms.

For our sequence, we have `a_i = 2i` for `i = 1, 2,...`

and we want to solve: `sum_(i=1)^(i=n) a_i = 60762` to find `n`

`60762 = sum_(i=1)^(i=n) a_i = n((a_1+a_n)/2) = n((2+2n)/2) = n(n+1)`

Now `n^2 < n(n+1) = 60762 < (n+1)^2`

So `n < sqrt(60762) < n+1`

So `n = floor(sqrt(60762)) = 246`

Check: `246 * 247 = 60762`