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The sum of three numbers in A.P. is $3$ and their product is $- 35$. Find the numbers.
Given: The sum of three numbers in A.P. is $3$ and their product is $-35$.
To do: To find the numbers.
Solution:
From the question it is given that,
The sum of three numbers in A.P. $=3$
Given, Their product $=-35$
Let us assume the $3$ numbers which are in A.P. are, $a-d,\ a,\ a+d$
Now adding $3$ numbers $=a-d+a+a+d=3$
$\Rightarrow 3a=3$
$\Rightarrow a=\frac{3}{3}$
$\Rightarrow a=1$
From the question, product of $3$ numbers is $-35$
So, $( a-d)\times( a)\times( a+d)=-35$
$\Rightarrow (1-d)\times(1)\times(1+d)=-35$
$\Rightarrow 1^2-d^2=-35$
$\Rightarrow d^2=35+1$
$\Rightarrow d^2=36$
$\Rightarrow d=\pm\sqrt{36}=\pm6$
If $d=6$, the numbers are:
$(a-d)=1-6=-5$
$a=1$
$(a+d)=1+6=7$
If $d=-6$, the numbers are $(a-d)=1-(-6)=1+6=7$
$a=1$
$(a+d)=1+(-6)=1-6=-5$
Therefore, the numbers $-5,\ 1,\ 7,\ …$ and $7,\ 1,\ -5,\ …$ are in A.P.
Updated on: 10-Oct-2022
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