I wanna figure out the numbers in an arithmetic sequence question

Question

12. The sum of three consecutive terms in an arithmetic sequence is 21 and the product of the two extreme numbers is 45. Find the numbers.

How would we go on about finding this?!

Answer 1

We have two equations. From the first sentence, we have $(x - d) + x + (x + d) = 21$, where $d$ is the common difference of the arithmetic sequence. We can solve this equation to find $x$, though $d$ is still unknown. Then from the second sequence, we have $(x - d)(x + d) = 45$; with the known value of $x$, we can solve for $d$, then use $x$ and $d$ to find the original three numbers.

Answer 2

Say we have a series $\dots ,a-x,a,a+x,\dots$. Then $$21 = (a-x) + a + (a+x) = 3a.$$ Hence, $a = 7$. As we also have $$45 = (a-x)(a+x) = (7-x)(7+x) = 49 - x^2$$. Hence, $x^2 = 4$. So we get $x = 2\vee x = -2$.

So the numbers are $(7,2)$ and $(7,-2)$.

Note that one gives us an increasing sequence, while the other gives a decreasing sequence.