Question

Find the four numbers in A.P, whose sum is $50$ and in which the greatest number is four times the least.

Answer 1

Let the four numbers be $a-3d,a-d,a+d,a+3d$

Now,

$a-3d+a-d+a+d+a+3d=50$

or, $4a=50\Rightarrow a=\frac{25}{2}$

Let $d$ be a positive integer, then greatest number and least number is $(a-3d)$

According to question, $a+3d=4(a-3d)$

or, $a+3d=4a-12d$

or, $a=5d$

So, $\frac{25}{2}=5d.\Rightarrow d=\frac{5}{2}$

$\therefore$ The number in $A.P$ are:

$(\frac{25}{2}-\frac{15}{2}),(\frac{25}{2}-\frac{5}{2}),(\frac{25}{2}+\frac{5}{2}),(\frac{25}{2}+\frac{15}{2})$

$=5,10,15,20$, when least number $5=4$ (greatest number)

$=4\times 5=20$