Arithmetic Sequences: Find three consecutive numbers where the product of the three number is 11 times larger than their sum.

Question

Hi thx for any help/ideas that anyone can offer.

This is an Arithmetic Sequences problem: Find three consecutive numbers where the product of the three integer number is 11 times larger than their sum. Find the three terms. So far I'm stuck at this point

       T(1) = a - d
       T(2) = a
       T(3) = a + d
       Sum      =  3a
       Product  =  (a)( a^2 - d^2)

let

       Sum x 11 = Product
       3a  x 11 = (a)(a^2 - d^2)

we have

       a^2 - d^2 = 33

I'm stuck at this point where the question doesn't have other information on their relationships to make another equation to solve for a or d.

By eleven times larger, does this mean twelve times? Also consecutive would mean differing by $1$? — Mark Bennet, Aug 12 '17 at 17:17
Note: I assumed that by "consecutive" you meant successive terms in an Arithmetic Progression with period $d$. You should clarify. — lulu, Aug 12 '17 at 17:18
There is an error (or more strictly oversight) in the proposed workings. — Mark Bennet, Aug 12 '17 at 17:20
Changed, my mind. Consecutive must must arithemetically consecutive with period d. But $a^2 - d^2 =(a+d)(a-d) = 33= 3*11$ so $a-d = 3,1,-11,-33$ and $a +d = 11, 33, -3, -1$ are clearly solvable (4 sets of solutions, if we assume $a \ne 0$ and infinite solutions iv we don't.). — fleablood, Aug 12 '17 at 17:24
As long as we are quibbling: the consecutive numbers does not nesc. mean integers. In which case $33 = a^2 - d^2$ has a continuum of solutions. I think we need clarification. My first blush impression is "consecutive" numbers is $k, k+1, k+2; k \in \mathbb Z$ in which case the only solution is $-1, 0, 1$ which I don't like because if I use coloquialisms for "consecutive numbers" I should also interpret "times larger" as purely positive in which there are no solutions. — fleablood, Aug 12 '17 at 17:48
Sorry, this is the first time I ever post math question online. Thanks for all the help, you guys are amazing. — Bluze, Aug 12 '17 at 18:48

score 1 · Answer 1 · answered Aug 12 '17 at 17:21

1

Supposing $1 \le d <a$ we have to solve $$(a+d)(a-d)=33$$ Since $33$ has only four divisors: $1,3,11,33$ we have $$\begin{cases} a+d=11 \\ a-d=3\end{cases} \ \ \mathrm{ or } \ \ \begin{cases} a+d=33 \\ a-d=1\end{cases}$$ which gives the two solutions $$\begin{cases} a=7 \\ d=4\end{cases} \ \ \mathrm{ or } \ \ \begin{cases} a=17 \\ d=16\end{cases}$$

answered Aug 12 '17 at 17:21

Crostul

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As mark bennet points out we should consider $a = 0$ and, I suppose we should also consider $a +d$ and $a -d$ both negative. – fleablood Aug 12 '17 at 17:26
thank you for putting it in a way i understand <3 – Bluze Aug 12 '17 at 18:49

score 0 · Answer 2 · answered Aug 12 '17 at 17:23

With the three numbers $a-1, a, a+1$ and the product $r$ times the sum, we have $$a^3-a=3ra$$ Which means $a^2-1=3r$

or $a=0$

This also works, of course, for arbitrary differences $d$. For some $d$ there may be additional integer solutions, obtained as others have suggested.

score 0 · Answer 3 · answered Aug 12 '17 at 17:26

$a^2-d^2=33$

so $(a+d)(a-d)=33$

now what is the number of the ways in 33 can be factorized in two integers.

$1*33$ or $11*3$ or $33*1$ or $3*11$.

so now $a+d=1$ and $a-d=33$ solve these eqautions for integers.

a=33 and d=0 but d can't be zero so solve for the next pair.

$a+d=3$ and $a-d=11$ and repeat this step for every pair!

fleablood · Accepted Answer · 2017-08-12T22:29:37.313

You derived

$33a = a(a^2 - d^2) = a(a-d)(a+d)$

But before we divide both sides by $a$, we must consider that $a$ may equal $0$.

So Case 1: If $a = 0$.

1) $a = 0; d\in \mathbb Z$. And $(-k, 0 , k)$ are solutions. (Trivial solutions)

Those are infinite number of solutions.

Cases 2-???: $a \ne 0$.

Then we can divide both sides by $a$.

2) $a \ne 0$ and $(a-d)(a+d) = 33$

So $a-d$ and $a+d$ are two factors of $33$. $33 = 3*11$ has as factors $\pm 1, \pm 3, \pm 11, \pm 33$.

So we have eight case all very similar:

a) $a + d = 1$ and $a-d = 33$. So $a= 17$ and $d= -16$ and the numbers are $1, 17, 33$.

b) $a + d = 33$ and $a - d = 1$. So $a = 17$ and $d = 16$ and the numbers are $1, 17, 33$.

and ...

this is too tedious to spell out. We know that the pair $(a+d)$ and $(a-d)$ will be one of these pairs $(1, 33) , (-1, -33), (3,33)(-3,-33)$ and $a$ will be the number in the middle. And we know that solving it for negative numbers will be the same as solving it for positive numbers.

So we have the numbers are:

$$1, 17, 33$$ $$-1, -17,-33$$ $$3,7, 11$$ $$-3, -7, -11$$.

=====

Note, if "consecutive" integers mean $k, k+1, k+2$ i.e. with period specifically $1$, which is often (but not always) implied. Then $(-1, 0 , 1)$ is the only solution.

I believe you got the correct answers, yet I don't quite get how you got it. But thank you for the time, appreciate it!! <3 — Bluze, Aug 12 '17 at 18:56
Really? I'll try to explain it better detail be it boils down to $(a-d)$ and $(a+d)$ are factors of 33, which has only 1,3,11, and 33 as factors. — fleablood, Aug 12 '17 at 22:11
Wow thanks for the detail explanation!! I now have a much better understanding of this question. Thank you thank you <3 — Bluze, Aug 13 '17 at 14:05

Arithmetic Sequences: Find three consecutive numbers where the product of the three number is 11 times larger than their sum.

4 Answers4