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Now I already know the answer to that you have to do this first

$3x=36$

And then the rest you are just subbing in, but why do you have to divide 36 by 3. I am confused.

MANI
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Exodus
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    You can represent the terms as $a-r$, $a$ and $a+r$. Then the sum is given by $3a= 36 \implies a = 12$. For the product, you have $(a-r) \times a \times (a+r) = 1428 \implies (a^2 - r^2) \times a = 1428$. Since the value of $a$ is now known, this can easily be solved. – Nash J. Jan 09 '20 at 12:41
  • Presumably $x$ is the middle term, i.e., $(x-d)+x+(x+d)=36$, so the rest is $(x-d)x(x+d)=1428$. The trick is to solve these two equations for the two unknowns, $x$ and $d$. – Barry Cipra Jan 09 '20 at 12:41
  • The first thing you should do is define $x$, then you should explain why $3x=36$. When you do that, dividing by $3$ is the way to solve your equation. – Ross Millikan Jan 09 '20 at 14:45

4 Answers4

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The sum of an arithmetic series with an odd number of terms

is the middle term times the number of terms:

$\Sigma = n \dfrac{a_1+a_n}2.$

In this example, $36=3x$.

J. W. Tanner
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The three terms can be written $x-d$, $x$, $x+d$. Then

$$(x-d)+x+(x+d) = 3x =36.$$

Since $3$ times $x$ is $36$, divide both sides by $3$ to get $x=12$. Now the three terms are $12-d$, $12$, and $12+d$. Multiply them together to get

$$(12-d)(12)(12+d) = 1428.$$

Divide both sides by $12$ and the remaining equation gives you $d$. Then you can get all three terms.

B. Goddard
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$$\begin{align} a+(a+d)+(a+2d)&=36\\ 3a+3d&=36\\ a+d&=12 \end{align}$$

saulspatz
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We can let the 3 numbers be $(x, x-y, x+y)$ if we take the sum we see

\begin{align} (x-y)+x+(x+y) &= 36\\3x&=36\\x&=12 \end{align} solving product equation gives you

\begin{align} (12-y)(12+y)*12&=1428\\(12^2 - y^2)&=119\\144-119&=y^2\\y=\pm5\end{align} we can use any value of y because of the way we have written the 3 numbers in the beginning.

Leafy
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