# Arithmetic Progression - two series

I have two arithmetic progressions: $a, b, c, d$ and $w, x, y, z$ If the arithmetic progressions are merged together like this: $aw, bx, cy, dz$, is it possible to find the sum of the series?

Let $a$ be the first term and $c$ be the last term of the series. Let $n$ be the number of terms in the series and $b$ the common difference.

$$\frac{\sin\frac{a + c}{3}\sin\frac{nb}{2}}{\sin{nb/2}}$$

• The sum of what series? – chubakueno Jul 1 '14 at 16:22
• It looks as if you are taking term by term products. Of course one can find the sum, just add. More generally, if you have two given arithmetic progressions of arbitrary length $n$, you can find a closed-form formula for the series of termwise products. – André Nicolas Jul 1 '14 at 16:22
• Note that $\sin 2, \sin 4, \sin 6 \dots$ is not an arithmetic progression. – Ross Millikan Jul 1 '14 at 16:27
• Consider that $\sin2=\sin178$. Sum your series $S$ with the reversed version of it and... – chubakueno Jul 1 '14 at 16:30
• Ohhhh!!!!!!!! I see, let me try that. – Gummy bears Jul 1 '14 at 16:32

The $i^{\text{th}}$ of the first series (if we start counting at zero) is $a+i(b-a)$. The $i^{\text{th}}$ term of your combined series is then $[a+i(b-a)][w+i(x-w)]=aw+i[a(x-w)+w(b-a)]+i^2(b-a)(x-w)$ Now you can use the sum of powers formulas to sum over the range of $i$ you desire.
• Yes, they are from your question. Because it is an arithmetic series, you must have $d=a+3(b-a)$, for example. – Ross Millikan Jul 1 '14 at 16:28
• It doesn't look complicated to me. Remember, for example, $\sum_{i=0}^n i=\frac 12n(n+1)$. All the other pieces are constants for this purpose. This gives you an explicit formula that only uses $a,b,w,x$ and the number of terms. – Ross Millikan Jul 1 '14 at 16:32
I have two arithmetic progressions: $a, b, c, d$ and $w, x, y, z$ If the arithmetic progressions are merged together like this: $aw, bx, cy, dz$, is it possible to find the sum of the series?
The result is $$(b+c)(x+y)+5(c-b)(y-x),$$ or, equivalently, $$2b(3x-2y)+2c(3y-2x),$$ or, equivalently, $$2x(3b-2c)+2y(3c-2b),$$ or, finally, $$6(bx+cy)-4(by+cx).$$ Proof: Use $a=2b-c$, $d=2c-b$, $w=2x-y$, $z=2y-x$, and simplify.