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

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Last updated date: 18th Jul 2024
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Answer
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Hint: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 \[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d,n=1,2,3....\]