Question

The sum of the three numbers in A.P is $21$ and the product of their extremes is $45$. Find the numbers

• A. $5,7$ and $9$
• B. $9,7$ and $5$
• C. $7,9$ and $11$
• D. Both $\left(1\right)$ and $\left(2\right)$
• E. None of these

Answer 1

Answer: (D)

Both $\left(1\right)$ and $\left(2\right)$

Let the three number be $a-d,a,a+d$
Given, $(a-d)+a+(a+d)=21\Rightarrow 3a=21\Rightarrow a=7$
Also, $(a-d)(a+d)=45$
$\Rightarrow (7-d)(7+d)=45$
$\Rightarrow (49-d^2)=45$
$\Rightarrow d^2=4$
$\Rightarrow d=-2,2$
Therefore, the numbers are:
when $d=2$; $7-2,7,7+2$i.e.$5,7,9$
when $d=-2$; $7+2,7,7-2$i.e.$9,7,5$

Hence, option C is correct.