If four numbers in A.P. are such that their sum is 50

Here, we are given that four numbers are in A.P., such that their sum is 50 and the greatest number is 4 times the smallest.

So, let us take the four terms as $a-d, a, a+d, a+2 d$.

Now, we are given that sum of these numbers is 50, so we get,

$(a-d)+(a)+(a+d)+(a+2 d)=50$

$a-d+a+a+d+a+2 d=50$

$4 a+2 d=50$

$2 a+d=25$ .........(1)

Also, the greatest number is 4 times the smallest, so we get,

$a+2 d=4(a-d)$

$a+2 d=4 a-4 d$

$4 d+2 d=4 a-a$

$6 d=3 a$

$d=\frac{3}{6} a$..........(2)

Now, using (2) in (1), we get,

$2 a+\frac{3}{6} a=25$

$\frac{12 a+3 a}{6}=25$

$15 a=150$

$a=\frac{150}{15}$

$a=10$

Now, using the value of a in (2), we get,

$d=\frac{3}{6}(10)$

$d=\frac{10}{2}$

$d=5$

So, first term is given by,

$a-d=10-5$

$=5$

Second term is given by,

$a=10$

Third term is given by,

$a+d=10+5$

$=15$

Fourth term is given by,

$a+2 d=10+(2)(5)$

$=10+10$

$=20$

Therefore, the four terms are $5,10,15,20$

Hence, the correct option is (a).