The sum of three numbers in an A.P is $-12$ and their product is $36$. Find the numbers.
Let the three numbers in A.P be $a - d, \; a, \; a + d$
where $a =$ first term of A.P; $\;$ $d =$ common difference of A.P
As per question,
$a - d + a + a + d = -12$
i.e. $\;$ $3a = -12$ $\implies$ $a = -4$
and $\;$ $\left(a - d\right) \times a \times \left(a + d\right) = 36$
i.e. $\;$ $a^2 - d^2 = \dfrac{36}{a}$
i.e. $\;$ $\left(-4\right)^2 - d^2 = \dfrac{36}{-4}$ $\;\;$ [substituting the value of $a$]
i.e. $\;$ $16 - d^2 = -9$
i.e. $\;$ $d^2 = 25$ $\implies$ $d = \pm 5$
When $d = + 5$, the three numbers are $-4 - 5, \; -4, \; -4 + 5$ $\;$ i.e. $\;$ $-9, \; -4, \; 1$
When $d = - 5$, the three numbers are $-4 + 5, \; -4, \; -4 - 5$ $\;$ i.e. $\;$ $1, \; -4, \; -9$