Termwise Product of Arithmetic Progressions

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Theorem


Let us consider two different arithmetic progression series.

\(a , a+d , a+2d , a+3d ... a+(n-1)d\)

\(a_1 , a_1+d_1 , a_1+2d_1 , a_1+3d_1 ... a_1+(n-1)d_1\)

Now the series formed by product of their respective terms will be

\(aa_1 ,(a+d)(a_1+d_1) ,(a+2d)(a_1+2d_1),(a+3d)(a_1+3d_1) ...[a+(n-1)] a_1+(n-1)d_1 \)

So the summation of all the terms will be

\(\Sigma a_1+(a+d)(a_1+d_1) ...+[a+(n-1)d][a_1+(n-1)d_1] \)

Giving the Termwise Product of Arithmetic Progressions

\(n(a_1a)+d_1d \frac{(n-1)(n)(2n-1)}{6}+[{a_1d+ad_1}]\frac{(n-1)n}2 \)

Examples

Let us take series of 1,2,3.....10

And the second series of 1,3,5.....19

Now sum of product of respective terms will be\[10(1 \times 1)+1 \times 2 \frac{(10-1)(10)(20-1)}{6}+[{2+1}]\frac{(10-1)10}2 = 430 \]

and this works for any arithmetic series in real numbers.

Derivation

Let the two arithmetic series be

\(a , a+d , a+2d , a+3d ... a+(n-1)d\)

\(a_1 , a_1+d_1 , a_1+2d_1 , a_1+3d_1 ... a_1+(n-1)d_1\)

\(aa_1 ,(a+d)(a_1+d_1) ,(a+2d)(a_1+2d_1),(a+3d)(a_1+3d_1) ...[a+(n-1)] a_1+(n-1)d_1 \)

and sum of all the terms of this third series will be

\(\Sigma a_1+(a+d)(a_1+d_1) ...+[a+(n-1)d][a_1+(n-1)d_1] \)

Now we can use foil method to open-up brackets and get each term.

\(\Sigma a_1a+a_1a+d_1d+a_1d+d_1a ...+a_1a+d_1d+a_1d+d_1a (n-1)^2 \)

Now using laws of algebra and formula of sum of squares of Arithmetic Series we get,

\(n(a_1a)+d_1d \frac{(n-1)(n)(2n-1)}{6}+[{a_1d+ad_1}]\frac{(n-1)n}2 \)