Question

The sum of three consecutive terms in an arithmetic progression is $6$ and their product is $-120$. Find the three numbers.

Answer 1

Let $(a+d),(a+2d)$ and $(a+3d)$ be three consecutive terms in A.P.

As given, $(a+d)+(a+2d)+(a+3d)=6$ ...... $\left(1\right)$

and $(a+d)(a+2d)(a+3d)=-120$ ....... $\left(2\right)$

From first equation $(a+d)+(a+2d)+(a+3d)=6$ we get,

$3a+6d=6$

$⟹$ $a+2d=2$

$\therefore$ $a=2-2d$

Substitute value of $a$ in the equation $\left(2\right)$ we get,

$(2-2d+d)(2-2d+2d)(2-2d+3d)=-120$

$⟹$ $(2-d)\left(2\right)(2+d)=-120$

$⟹$ $(4-d^2)=-60$

$⟹$ $d^2=64$

$\therefore$ $d=\pm 8$

Hence, there are two possibilities

(i) If $d=8$ then $a=2-2\times 8=-14$

Then the three consecutive numbers would be $-6,2,10$

(ii) If $d=-8$ then $a=2-2\times (-8)=18$

Then three consecutive numbers would be $10,2,-6$

$\therefore$ The three consecutive numbers are $-6,2,10$