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

Jul 14, 2015

Let ${a}_{i} = 2 i$ for $i = 1 , 2 , \ldots$

$60762 = {\sum}_{i = 1}^{i = n} {a}_{i} = n \left(\frac{{a}_{1} + {a}_{n}}{2}\right) = n \left(\frac{2 + 2 n}{2}\right) = n \left(n + 1\right)$

So $n = \left\lfloor \sqrt{60762} \right\rfloor = 246$

#### Explanation:

A standard property of sums of arithmetic sequences is:

${\sum}_{i = 1}^{i = n} {a}_{i} = n \left(\frac{{a}_{1} + {a}_{n}}{2}\right)$

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 ${a}_{i} = 2 i$ for $i = 1 , 2 , \ldots$

and we want to solve: ${\sum}_{i = 1}^{i = n} {a}_{i} = 60762$ to find $n$

$60762 = {\sum}_{i = 1}^{i = n} {a}_{i} = n \left(\frac{{a}_{1} + {a}_{n}}{2}\right) = n \left(\frac{2 + 2 n}{2}\right) = n \left(n + 1\right)$

Now ${n}^{2} < n \left(n + 1\right) = 60762 < {\left(n + 1\right)}^{2}$

So $n < \sqrt{60762} < n + 1$

So $n = \left\lfloor \sqrt{60762} \right\rfloor = 246$

Check: $246 \cdot 247 = 60762$