Question 1206662
<pre>
If you knew nothing about arithmetic sequences, you would add this row
of 10 numbers:

15 + 17 + 19 + 21 + 23 + 25 + 27 + 29 + 31 + 33

But since you know about arithmetic sequences you can substitute in

{{{S[n]}}}{{{""=""}}}{{{expr(n/2)(2a[1]+(n-1)d)}}}

with a<sub>1</sub> = 15, d = 2, and n = 10

{{{S[10]}}}{{{""=""}}}{{{expr(10/2)(2(15)^""+(10-1)(2))}}}
{{{S[10]}}}{{{""=""}}}{{{5(30+(9)(2)^"")}}}
{{{S[10]}}}{{{""=""}}}{{{5(30+18^"")}}}
{{{S[10]}}}{{{""=""}}}{{{5(48^"")}}}
{{{S[10]}}}{{{""=""}}}{{{240^""}}}

The formula saved you from having to add 10 numbers together.

It wasn't that great a help here, but some problems will require
you to add 100 or even 1000 numbers. Then it will be a much bigger
help.  In school you are often given easier problems to show you
the principle, so you can use it on harder problems.

Edwin</pre>