Three consecutive terms of an arithmetic sequence have a sum of 36 and a product of 1428. Find three terms.

Question

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.

Answer 1

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

Answer 2

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, $36=3x$.

Answer 3

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.

Answer 4

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.