Question

Find the four numbers in AP whose sum is 50 and in which the greatest number is 4 times the least .

Answer 1

Let the 4 numbers be
$a-3d,a-d,a+d,a+3d$
then
$\Rightarrow$ $a-3d+a-d+a+d+a+3d=50$
$\Rightarrow$ $4a=50$
$\therefore$ $a=\frac{25}{2}$

And,
$\Rightarrow$ $a+3d=4(a-3d)$
$\Rightarrow$ $a+3d=4a-12d$
$\Rightarrow$ $3a=15d$
$\Rightarrow$ $a=5d$
$\Rightarrow$ $d=\frac{a}{5}$

$\therefore$ $d=\frac{25}{10}$

So the four numbers are
$a-3d=\frac{25}{2}-3\times \frac{25}{10}=5$

$a-d=\frac{25}{2}-\frac{25}{10}=10$

$a+d=\frac{25}{2}+\frac{25}{10}=15$

$a+3d=\frac{25}{2}+3\times \frac{25}{10}=20$

Hence, four numbers are $5,10,15,20$ .