# Thread: Multiplication of arithmetic series to get 3rd series w/ 1st 3 terms: 468, 462, 384

1. ## Multiplication of arithmetic series to get 3rd series w/ 1st 3 terms: 468, 462, 384

Hello All

I am stuck on this problem

Two arithmetic sequences are multiplied together term by term to form another sequence whose first three terms are 468, 462 and 384. What is the fourth term?

I can't see a way into it without resorting to heavy algebra or making some guesses?

2. Can you write a general expression for the terms of an arithmetic series?

3. Originally Posted by tkhunny
Can you write a general expression for the terms of an arithmetic series?
yes a+(n-1)d

so i get equations like ab= 468
(a+d)(b+e) = 462
(a+2d)(b+2e) = 324

But then what..?

4. Originally Posted by Denis
Make up and play around with a short simple example, like:
2 5 8 11
3 7 11 15
------------
6 35 88 165
Should i just then work out the nth term of a quadratic sequence? ( generated by multiplying the two linear ones)

5. Originally Posted by apple2357
yes a+(n-1)d

so i get equations like ab= 468
(a+d)(b+e) = 462
(a+2d)(b+2e) = 384

But then what..?
See where it leads. Multiply stuff out:

. . . . .i. ab = 468

. . . . .ii. ab + bd + ae + de = 462

. . . . .iii. ab + 2bd + 2ae + 4de = 384

Subtracting (i) from each of (ii) and (iii) seems an obvious move, so let's see where that goes:

. . . . .iv. bd + ae + de = -6

. . . . .v. 2bd + 2ae + 4de = -84

The second new equation has all-even coefficients, so do the obvious simplification:

. . . . .vi. bd + ae + 2de = -42

Then comes an obvious subtraction, and so forth. Just try stuff, and see what works!

6. I see thanks. I will play with that.
I am wondering though, i will never be able to solve for the variables as there are 4 unknowns and only 3 equations?

7. Originally Posted by apple2357
I see thanks. I will play with that.I am wondering though, i will never be able to solve for the variables as there are 4 unknowns and only 3 equations?
Maybe you don't need to find the 4 unknowns. You are not asked to determine the sequences.

8. Originally Posted by stapel
See where it leads. Multiply stuff out:

. . . . .i. ab = 468

. . . . .ii. ab + bd + ae + de = 462

. . . . .iii. ab + 2bd + 2ae + 4de = 324 should be 384

Subtracting (i) from each of (ii) and (iii) seems an obvious move, so let's see where that goes:

. . . . .iv. bd + ae + de = -6

. . . . .v. 2bd + 2ae + 4de = -144, so this becomes -84

The second new equation has all-even coefficients, so do the obvious simplification:

. . . . .vi. bd + ae + 2de = -72, so this becomes -42

Then comes an obvious subtraction, and so forth. Just try stuff, and see what works!

My mistake above i miscopied 384 as 324, doesn't really matter...

But i think i have it.

Following your reasoning de = -36

then (bd+ae)= 30

I am now after (a+3d)(b+3e) = 468+ 3x(30) -9x36 = 234

Is this correct?

in messing with the algebra, i have discovered

u_4 = u_1 + 3u_2-3u_3

where u_1, u_2 and u_3 are any 3 terms of the sequence? I think

Thanks for help all!

9. Now, is it possible to actually work out the two original sequences?

10. Originally Posted by apple2357
Now, is it possible to actually work out the two original sequences?
Let's say first sequence (A) was obtained by multiplying another sequence (C) by a constant N. So we are looking at the product of sequences A and B, or C*N and B. Now let's look at the product of C and B*N. The product will be the same, but the original sequence are different. So, in general, I'd say you can't work it out. If one of the given products had very few factors...

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