Question

The sum of three numbers in A.P is $27$, and their product is $504$; find them.

Answer 1

Hint: In this question we have been given that the sum of three numbers is $27$ and their product is $504$ and the given numbers are in arithmetic progression. We will consider the numbers in arithmetic progression as $a-d,a,a+d$ where $d$ represents the common difference. We will make equations based on the given data and solve for the values of $a$ and $d$, and then substitute to get the required numbers.

Complete step by step solution:
Consider the three numbers to be $a-d,a,a+d$
Now the addition of the $3$ terms is $27$ therefore, we can write:
$\left( a-d \right)+\left( a \right)+\left( a+d \right)=27\to \left( 1 \right)$
Now the product of the $3$ terms is $504$ therefore, we can write:
$\left( a-d \right)\left( a \right)\left( a+d \right)=504\to \left( 2 \right)$
Now consider equation $\left( 1 \right)$
$\Rightarrow \left( a-d \right)+\left( a \right)+\left( a+d \right)=27$
On opening the brackets, we get:
$\Rightarrow a-d+a+a+d=27$
On simplifying, we get:
$\Rightarrow 3a=27$
On transferring the term $3$ from the left-hand side to the right-hand side, we get:
$\Rightarrow a=\dfrac{27}{3}$
On simplifying, we get:
$\Rightarrow a=9$, which is the value of $a$.
On simplifying the value of $a=9$ in equation $\left( 2 \right)$, we get:
$\Rightarrow \left( 9-d \right)\left( 9 \right)\left( 9+d \right)=504$
We can write it as:
$\Rightarrow \left( {{9}^{2}}-{{d}^{2}} \right)\left( 9 \right)=504$
On dividing both the sides by $9$, we get:
$\Rightarrow {{9}^{2}}-{{d}^{2}}=54$
On simplifying, we get:
$\Rightarrow {{d}^{2}}=81-54$
On subtracting, we get:
$\Rightarrow {{d}^{2}}=25$
On taking the square root on both the sides, we get:
$\Rightarrow d=\pm 5$, which is the value of $d$.
If $d=5$, the series, will be:
$\Rightarrow \left( 9-5 \right),9,\left( 9+5 \right)=4,9,14$
If $d=-5$, the series, will be:
$\Rightarrow \left( 9-\left( -5 \right) \right),9,\left( 9+\left( -5 \right) \right)=14,9,4$
Since in both cases the numbers are the same the numbers are $4,9,14$ which is the required solution.

Note: The general formula of arithmetic progression should be remembered which is ${{a}_{n}}=a+\left( n-1 \right)\times d$ where $a$ is the first term, $d$ is the common difference, $n$ is the number of terms and ${{a}_{n}}$ is the ${{n}^{th}}$ term in the arithmetic progression.