Question 968156
Find the indicated sum:

6 + 12 + 18 + ... + 300
<pre>
One of two formula for the sum of n terms is

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

We have a<sub>1</sub> = 6 and a<sub>n</sub> = 300

so we must find d and n

6 + 12 + 18 + ... + 300

First we determine the common difference, d

d = (2nd term)-(1st term) = (3rd term)-(2nd term) = 12-6 = 18-12 = 6

Next we determine the number of terms, n:

We use the formula for the nth term

{{{a[n]}}}{{{""=""}}}{{{a[1]+(n-1)d}}}

{{{300}}}{{{""=""}}}{{{6+(n-1)6}}}

{{{300}}}{{{""=""}}}{{{6+6(n-1)}}}

{{{300}}}{{{""=""}}}{{{6+6n-6}}}

{{{300}}}{{{""=""}}}{{{6n}}}

{{{50}}}{{{""=""}}}{{{n}}}

Now we can substitute in

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

{{{S[50]}}}{{{""=""}}}{{{expr(50/2)(6+300)}}}

{{{S[50]}}}{{{""=""}}}{{{25(306)}}}

{{{S[50]}}}{{{""=""}}}{{{7650}}}

---------------

Notice that we could have used the other sum formula just as well:

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

{{{S[50]}}}{{{""=""}}}{{{expr(50/2)(2(6)^""+(50-1)6)}}}

{{{S[50]}}}{{{""=""}}}{{{25(12^""+(49)6)}}}

{{{S[50]}}}{{{""=""}}}{{{25(12+294)}}}

{{{S[50]}}}{{{""=""}}}{{{25(306)}}}

{{{S[50]}}}{{{""=""}}}{{{7650}}}

Edwin</pre>