If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are

Answer: Option A

4 numbers are in A.P.
Let the numbers be
a – 3d, a – d, a + d, a + 3d
Where a is the first term and 2d is the common difference
Now their sum = 50
a – 3d + a – d + a + d + a + 3d = 50
and greatest number is 4 times the least number
a + 3d = 4 (a – 3d)
a + 3d = 4a – 12d
4a – a = 3d + 12d
⇒ 3a = 15d
$$\begin{array}{rl}& \Rightarrow a=\frac{15d}{3}=5d\\ & \Rightarrow \frac{25}{2}=5d\\ & \Rightarrow d=\frac{25}{2\times 5}\\ & \Rightarrow d=\frac{5}{2}\\ & \therefore \text{Numbers}\text{are}\end{array}$$
$$\frac{25}{2}-3\times \frac{5}{2},$$   $$\frac{25}{2}-\frac{5}{2},$$   $$\frac{25}{2}+\frac{5}{2},$$   $$\frac{25}{2}+3\times \frac{5}{2}$$
$$\begin{array}{rl}& \Rightarrow \frac{10}{2},\frac{20}{2},\frac{30}{2},\frac{40}{2}\\ & \Rightarrow 5,10,15,20\end{array}$$