Q.Three numbers are in Arithmetic Progression. Their sum is $15$ and their product is $80$. Find the numbers.

Take the three AP terms as $a-d,\ a,\ a+d$ so their sum gives $a$ at once.

$5(25 - d^2) = 80 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3$; the numbers are $2, 5, 8$.

Choosing the symmetric form $a - d,\ a,\ a + d$ for three terms in AP makes the sum simple.

Sum: $(a - d) + a + (a + d) = 3a = 15 \Rightarrow a = 5$.

Product: $(a - d)\,a\,(a + d) = a(a^2 - d^2) = 80$. Substituting $a = 5$:

With $a = 5,\ d = 3$ the numbers are $2, 5, 8$; with $d = -3$ they are $8, 5, 2$ — the same three numbers in reverse.

Independent check: $2 + 5 + 8 = 15$ ✓ and $2 \times 5 \times 8 = 80$ ✓.

Without the symmetric substitution one could write the terms as $a, a+d, a+2d$: sum $3a + 3d = 15 \Rightarrow a + d = 5$ (so the middle term is $5$), then the product condition again yields $d = \pm 3$ and the numbers $2, 5, 8$ — more algebra, same result. The symmetric form is preferred precisely because it makes the sum collapse to $3a$.

Taking the terms as $a, a+d, a+2d$ and then forgetting that the middle term equals the mean of the three (here $15/3 = 5$) — the symmetric $a-d, a, a+d$ form builds that fact in and avoids a messy simultaneous solve.