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

1 Answer
Jul 14, 2015

Answer:

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#