There are three numbers in A.P. whose sum is 24 and product is 480. Find the numbers.
1,5,10
6,8,10
2,5,7
1,5,10
Let the three numbers be a-d, a and a+d.
$\begin{aligned} & a-d+a+a+d=24 \\ & \Rightarrow 3 a=24 \\ & \Rightarrow a=8 \\ & (8-d)(8)(8+d)=480\end{aligned}$
$\begin{aligned} & \Rightarrow 64-d^2=60 \\ & \Rightarrow d^2=4 \\ & \Rightarrow d= \pm 2\end{aligned}$
Hence the three numbers are 6, 8 and 10 when d = 2 and the same numbers in reverse order, when$d=-2$.