Question

Three numbers are in AP such that their sum is $18$ and sum of their squares is $158$.

The greatest number among them is

• A. $10$
• B. $11$
• C. $12$
• D. None of these

Answer 1

The correct option is B

$11$

Explanation for the correct answer:

Let the three numbers in arithmetic progression be

$a-d,a,a+d$

It is given that the sum of these numbers is $18$

$\Rightarrow$ $a-d+a+a+d=18$

$\Rightarrow$ $3a=18$

$\Rightarrow$ $a=6$

It is given that the sum of their squares is $158$

$\Rightarrow$ ${\left(a-d\right)}^2+a^2+{\left(a+d\right)}^2=158$

Substitute value of $a=6$, we get

$\Rightarrow$ ${\left(6-d\right)}^2+6^2+{\left(6+\mathrm{d}\right)}^2=158$

$\Rightarrow$ $36-12d+d^2+36+36+12d+d^2=158$

$\Rightarrow$ $108+2d^2=158$

$\Rightarrow$ $2d^2=50$

$\Rightarrow$ $d^2=25$

$\Rightarrow$ $d=\pm 5$

Hence, the terms are $a-d=6-5=1$ or $a-d=6+5=11$

$a+d=6+5=11$ or $a+d=6-5=1$

Hence, the arithmetic progression is given as $1,6,11$ or $11,6,1$

Hence, the largest term for these arithmetic progressions is $11$.

Hence, option (B) is the correct answer.