Question

If sum 3 numbers in an AP is 24 and their product is 440 then find the numbers?

Answer 1

Let $a-d,a,a+d$ be the three terms of an A.P

Sum$=a-d+a+a+d=24$

or $3a=24\Rightarrow a=\frac{24}{3}=8$

Product $(a-d)a(a+d)=440$

$\Rightarrow (8-d)8(8+d)=440$

$\Rightarrow 64-d^2=55$

$\Rightarrow d^2=64-55=9$

$\Rightarrow d=\pm 3$

For $a=8,d=3$ the terms are $8-3,8,8+3$ or $5,8,11$

For $a=8,d=-3$ the terms are $8-(-3),8,8-3$ or $11,8,5$

$\therefore$ the numbers are $5,8,11$ and $11,8,5$