Question

The sum of three numbers in A.P. is $3$ and their product is $-35$. Find the numbers.

Answer 1

To find the three numbers that are in arithmetic progression, we take the number progression terms as $a-d,a,a+d$ first we will form an equation where we add these three numbers and equate it with the sum of the three numbers i.e. $3$ and after this we will find the product of the three terms and then equate it with $-35$.
For Sum: $\left( a-d \right)+a+\left( a+d \right)=3$
For Product: $\left( a-d \right)\times a\times \left( a+d \right)=-35$
As $a$ is the first term and $d$ is the difference value in each A.P.

Complete step by step solution:
According to the question given, we can say that the sum of the three numbers of an arithmetic progression is $3$ and the product of these three numbers are given as $-35$.
Now write the progression in terms of the first term and the middle term as $a$ is the first term and $d$ is the difference in each term. We get the value of progression as:
$\Rightarrow a-d,a,a+d$
After this we add the three numbers and make it equal to $3$ as given below:
$\Rightarrow \left( a-d \right)+a+\left( a+d \right)=3$
$\Rightarrow a+a+a=3$
$\Rightarrow a=1$
Similarly, as we have added the numbers of the arithmetic progression we will multiply the same and equate it with the value of $-35$ as given below:
$\Rightarrow \left( a-d \right)\times a\times \left( a+d \right)=-35$
Placing the value of $a=1$, we get the value of $d$ as:
$\Rightarrow 1-d\times 1\times 1+d=-35$
$\Rightarrow {{1}^{2}}-{{d}^{2}}=-35$
Changing the negative sign by interchanging the values from LHS to RHS as:
$\Rightarrow {{d}^{2}}=36$
$\Rightarrow d=\pm 6$
Now that we have got the numbers or the value of the first and the last term, we can place those values in the arithmetic terms $a-d,a,a+d$, we can get the value of the three numbers:
The value of the first number is given as:
$\Rightarrow a-d$
Placing the values in the above term for $a=1$ and $d=6$ (we can take either $6$ or $-6$), we get:
$\Rightarrow 1-6$
$\Rightarrow -5$
The value of the first number is given as:
$\Rightarrow a$
Placing the values in the above term for $a=1$ and $d=6$ (we can take either $6$ or $-6$), we get:
$\Rightarrow 1$
The value of the first number is given as:
$\Rightarrow a+d$
Placing the values in the above term for $a=1$ and $d=6$ (we can take either $6$ or $-6$), we get:
$\Rightarrow 1+6$
$\Rightarrow 7$
Therefore, the numbers that are given in arithmetic progression are given as: $-5,1,7$.

Note: Arithmetic progression is a method in which the sequence follows a constant difference pattern amongst the consecutive terms given. The formula for the nth term of a sequence is given as ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d,n=1,2,3....$