Question

Three numbers are in A.P. whose sum is $33$ and product is $792$, then the smallest number from these numbers is

• A. $4$
• B. $8$
• C. $11$
• D. $14$

Answer 1

The correct option is A

$4$

Explanation for the correct answer:

Step 1 : Assume suitable consecutive terms in arithmetic progression

Let the three consecutive terms in an arithmetic progression be

$a-d,a,a+d$

Step 2:Using given data find the value of $a$ and $d$

It is given that the sum of these numbers is $33$

$\Rightarrow$ $a-d+a+a+d=33$

$\Rightarrow$ $3a=33$

$\Rightarrow$ $a=11$

It is given that the product of these numbers is $792$

$\Rightarrow$ $\left(a-d\right)a\left(a+d\right)=792$

Substitute the value of $a=11$, we get

$\Rightarrow$ $11\left(11-d\right)\left(11+d\right)=792$

$\Rightarrow$ $121-d^2=72$

$\Rightarrow$ $d^2=49$

$\Rightarrow$ $d=\pm 7$

Step 3: Substitute the values of $a$ and $d$ to find the required terms

Hence, the terms of the arithmetic progression are

$a-d=11-7=4$ or $a-d=11-\left(-7\right)=18$

$a+d=11+7=18$ or $a+d=11+\left(-7\right)=4$

Hence, the arithmetic progression is either $4,11,18$ or $18,11,4$

Hence, the smallest term of this arithmetic progression is $4$.

Hence, option (A) is the correct answer.