Question

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

Answer 1

Let $a-d,a,a+d$ be the first three terms of an A.P.

It is given that the sum of the terms is $6$ that is:

$a-d+a+a+d=6\Rightarrow 3a=6\Rightarrow a=\frac{6}{3}\Rightarrow a=\mathrm{2......}\left(1\right)$

It is also given that the product of the terms is $-120$ that is:

$(a-d)\left(a\right)(a+d)=-120\Rightarrow a(a^2-d^2)=-120(\because (x+y\left)\right(x-y)=x^2-y^2)\Rightarrow 2(2^2-d^2)=-120\left(From\right(1\left)\right)\Rightarrow 4-d^2=-\frac{120}{2}$

$\Rightarrow 4-d^2=-60\Rightarrow d^2=4+60\Rightarrow d^2=64\Rightarrow d=\pm \sqrt{64}\Rightarrow d=\pm \mathrm{8...........}\left(2\right)$

Now, if $a=2$ and $d=-8$ then the first three terms of the A.P are:

$a-d=2-(-8)=2+8=10$

$a=2$ and

$a+d=2-8=-6$

And if $a=2$ and $d=8$ then the first three terms of the A.P are:

$a-d=2-8=-6$

$a=2$ and

$a+d=2+8=10$

Hence, the three terms are $-6,2,10$ or $10,2,-6$