Question

Find three numbers in A.P. whose sum and products are $21$ and $231$

respectively

Answer 1

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

It is given that the sum of the numbers is $21$, therefore,

$a-d+a+a+d=21\Rightarrow 3a=21\Rightarrow a=\frac{21}{3}=7......\left(1\right)$

It is also given that the product of the numbers is $231$, therefore using equation 1 we have,

$(a-d)\left(a\right)(a+d)=231\Rightarrow (7-d)\left(7\right)(7+d)=231\Rightarrow (7-d)(7+d)=\frac{231}{7}\Rightarrow 7^2-d^2=33\Rightarrow d^2=49-33\Rightarrow d^2=16\Rightarrow d=-4,d=4$

With $a=7$ and $d=4$, the three numbers are as follows:

$a-d=7-4=3$

$a=7$

$a+d=7+4=11$

Thus, the numbers are $3,7$ and $11$.

With $a=7$ and $d=-4$, the three numbers are:

$a-d=7-(-4)=7+4=11$

$a=7$

$a+d=7-4=3$

Thus, the numbers are $11,7$ and $3$.