<PRE><B>Question 1097076</B>
&lt;br&gt;Both of the responses you have received so far plug numbers into formulas to get the answer.  I believe students enjoy mathematics more if they understand what they are doing, rather than just plugging numbers into magic formulas.&lt;br&gt;
So let&#39;s look at this problem differently.&lt;br&gt;
The second tutor uses this formula for the sum of the terms of an arithmetic sequence, which I think is probably in nearly all algebra textbooks:
{{{S(n) = (n/2)(2a(1)+(n-1)d)}}}&lt;br&gt;
Let&#39;s try to understand what that formula means.&lt;br&gt;
First of all, (n/2) is clearly half the number of terms.&lt;br&gt;
Then what about that ugly expression in the next set of parentheses?&lt;br&gt;
Well, a(1) is the first term of the sequence; and a(1)+(n-1)d is the first term, a(1), plus the common difference, d, added (n-1) times.  And that is the formula for the last (n-th) term in the sequence.  So that ugly expression in the last set of parentheses is the sum of the first and last terms of the sequence.&lt;br&gt;
So this magic formula says that the sum of the terms of an arithmetic sequence is
(half the number of terms) times (the sum of the first and last terms)&lt;br&gt;
So why does that work?&lt;br&gt;
The key to understanding is to recognize that, because the terms in an arithmetic sequence are equally spaced, the sum of the first and last terms is the same as the sum of the second and next-to-last terms, and so on.  So this formula is grouping the n terms of the sequence into (n/2) pairs, each of which has the same sum as the first and last terms.&lt;br&gt;
So if you want to use that formula, don&#39;t memorize the formula without understanding it.  Instead, think what the formula says:&lt;br&gt;
sum = (half the number of terms) times (the sum of the first and last terms)&lt;br&gt;
That business about grouping the terms into pairs is fine; but I find it easier to think about the problem a bit differently.&lt;br&gt;
Instead of grouping the terms into (n/2) pairs each with the same sum as the first and last, I think of all of the n terms, multiplied by the AVERAGE of the first and last terms.&lt;br&gt;
So the formula the way I like to think of it is&lt;br&gt;
{{{S(n) = (n)((2a(1)+(n-1)d)/2)}}}&lt;br&gt;
But in fact I don&#39;t use that ugly formula; I simply think&lt;br&gt;
sum = (number of terms) times (average of first and last)&lt;br&gt;
So now, after all the discussion, here is how I would work your problem, using an understanding of what I am doing instead of magic formulas.&lt;br&gt;
My formula, again, is&lt;br&gt;
sum = (number of terms) times (average of first and last)&lt;br&gt;
I know the number of terms (30), and I know the first term (30,000); but I don&#39;t know the last term.&lt;br&gt;
But I know the common difference (1000), and a basic understanding of what an arithmetic sequence is tells me that the last (30th) term is the first term, plus the common difference 29 times:
last term = 30,000 + 29(1000) = 59,000&lt;br&gt;
Now I have what I need to find the answer.&lt;br&gt;
The average salary is the average of the first and last:
{{{(30000+59000)/2 = 44500}}}&lt;br&gt;
and so the sum of all the terms is
(number of terms) times (average):
{{{30*44500 = 1335000}}}&lt;br&gt;
&lt;br&gt;
I hope this response has helped you to a better understanding of how the sum of the terms of an arithmetic sequence is calculated; and that that better understanding helps make this easier for you.</PRE>