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. $noneofthese$

Answer 1

The correct option is B $11$

Let the nos be , $a-d,a,a+d$ then $a-d+a+a+d=18$

$3a=18\Rightarrow a=6$

$\therefore 6-d,6,6+d$

Sum of square = 158

$(6-d{)}^2+6^2+(6+d{)}^2=158$

$6^2+d^2-12d+6^2+6^2+d^2+12d=158$

$108+2d^2=158$

$2d^2=50$

$d^2=25\Rightarrow d=\pm 5$

when $d=+5$, Numbers are : 1, 6, 11
when $d=-5$, Numbers are : 11, 6, 1

Greatest : 11