Question

Find three numbers in A.P. whose sum and products are respectively $36$ and $1620$

Answer 1

Let the three numbers be $a-d,a,a+d$.

It is given that the sum of the numbers is $36$, therefore,

$a-d+a+a+d=36\Rightarrow 3a=36\Rightarrow a=\frac{36}{3}=12......\left(1\right)$

It is also given that the product of the numbers is $1620$, therefore using equation 1 we have,

$(a-d)\left(a\right)(a+d)=1620\Rightarrow (12-d)\left(12\right)(12+d)=1620\Rightarrow (12-d)(12+d)=\frac{1620}{12}\Rightarrow {12}^2-d^2=135\Rightarrow d^2=144-135\Rightarrow d^2=9\Rightarrow d=-3,d=3$

With $a=12$ and $d=3$, the three numbers are as follows:

$a-d=12-3=9$

$a=12$

$a+d=12+3=15$

Thus, the numbers are $9,12$ and $15$.

With $a=12$ and $d=-3$, the three numbers are:

$a-d=12-(-3)=12+3=15$

$a=12$

$a+d=12-3=9$

Thus, the numbers are $15,12$ and $9$.