# Arithmetic Sequence

• December 13th 2009, 04:50 AM
p4pri
Arithmetic Sequence
The rth term of an arithmetic progression is 1+4r. Find, in terms of n, the sum of the first n terms of the progression?

Thanks
• December 13th 2009, 06:10 AM
Hello p4pri
Quote:

Originally Posted by p4pri
The rth term of an arithmetic progression is 1+4r. Find, in terms of n, the sum of the first n terms of the progression?

Thanks

When $r=1$ we get the first term, which is $1+4=5$. Denote this by $a$.

When $r= 2$, we get the second term, which is $1 + 8 = 9$.

So the common difference is $9 - 5 = 4$. Denote this by $d$.

Use the formula for the sum of the first $n$ terms:
$S_n = \tfrac{1}{2}n(2a + [n-1]d)$

Can you complete this now?

• December 13th 2009, 07:29 AM
HallsofIvy
Arithmetic sequences have the very nice property that the average of all numbers in the sequence is the same as the average of the first and last numbers!

For example, the average value in the sequence 4, 7, 10, 13, 16 (an arithmetic sequence with first number 4 and "common difference 3) is (4+ 7+ 10+ 13+ 16)/5= 50/5= 10 (the middle number in the sequence of course) and (4+ 16)/2= 20/2= 10.

Notice that Grandad's formula, $S_n = \tfrac{1}{2}n(2a + [n-1]d)$ is, exactly, n times $\frac{a+ a+(n-1)d}{2}$, the average of the first and last terms.
• December 13th 2009, 07:56 AM
... and it's very easy to see why: just write the sequence down twice, one underneath the other - but the second one in reverse order. Then add each term to the one underneath it. You get the same total $n$ times over. Multiplying this total by $n$, and dividing the result by $2$, will then give the sum of the sequence, of course.