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?

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- Dec 13th 2009, 04:50 AMp4priArithmetic 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 - Dec 13th 2009, 06:10 AMGrandad
Hello p4priWhen $\displaystyle r=1$ we get the first term, which is $\displaystyle 1+4=5$. Denote this by $\displaystyle a$.

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

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

Use the formula for the sum of the first $\displaystyle n$ terms:$\displaystyle S_n = \tfrac{1}{2}n(2a + [n-1]d)$and there's your answer.

Can you complete this now?

Grandad - Dec 13th 2009, 07:29 AMHallsofIvy
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, $\displaystyle S_n = \tfrac{1}{2}n(2a + [n-1]d)$ is, exactly, n times $\displaystyle \frac{a+ a+(n-1)d}{2}$, the average of the first and last terms. - Dec 13th 2009, 07:56 AMGrandad
... 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 $\displaystyle n$ times over. Multiplying this total by $\displaystyle n$, and dividing the result by $\displaystyle 2$, will then give the sum of the sequence, of course.

Grandad