Browse Questions

Find three numbers in AP whose sum is 33 and the sum of whose squares is 395

$\begin{array}{1 1} 9,11,13 \\ 8,11,14 \\ 7,11,15 \\ 6,11,16 \end{array}$

Explanation : Let the numbers in AP be a-d,a,a+d
$sum=33=a-d+a+a+d$
$33=3a$
$a=11$
sum of squares =395
$=(a-d)^2+a^2+(a+d)^2=a^2-2ad+d^2+a^2+a^2+2ad+d^2$
$3a^2+0+2d^2=395$
$a=11\;,a^2=121$
$ad^2=395-363$
$d^2=\frac{32}{2}=16$
$d=4$
Numbers in AP are 7,11,15 .