Question

Find the three numbers in A.P. such that their sum is $24$ & the sum of their square is $200$.

Answer 1

Given: the number are in $A.P$

let the number be,

$(a-d),\left(a\right),(a+d)$

common difference $=d$

Sum $=24$

$\Rightarrow (a-d)+a+a+d=2a$

$\Rightarrow 3a=2a\Rightarrow a=8$

$\therefore$ the number are

$(8-d),8,(8+d)$

Sum of square $=200$

$\Rightarrow {(8-d)}^2+8^2+{(8+d)}^2=200$

$\Rightarrow 8^2+d^2-2\times 8\times d+8^2+8^2+d^2+2\times 8\times d=200$

$\Rightarrow 3\times \left(64\right)+2\left(d^2\right)=200$

$\Rightarrow 182+2d^2=200$

$\Rightarrow 2d^2=18\Rightarrow d^2=9\Rightarrow d=3$

$\therefore$ The three numbers are $8-3,8,8+3$

$\Rightarrow (5,8,11)$

if $d=-3$

the $3$ numbers will be

$8+3,8,8-3$

$\Rightarrow (11,8,5)$