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

Solution (By Examveda Team)

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
$$\eqalign{ & \Rightarrow a = \frac{{15d}}{3} = 5d \cr & \Rightarrow \frac{{25}}{2} = 5d \cr & \Rightarrow d = \frac{{25}}{{2 \times 5}} \cr & \Rightarrow d = \frac{5}{2} \cr & \therefore {\text{Numbers}}\,{\text{are}} \cr} $$
$$\frac{{25}}{2} - 3 \times \frac{5}{2},$$   $$\,\frac{{25}}{2} - \frac{5}{2},$$   $$\frac{{25}}{2} + \frac{5}{2},$$   $$\frac{{25}}{2} + 3 \times \frac{5}{2}$$
$$\eqalign{ & \Rightarrow \frac{{10}}{2},\,\frac{{20}}{2},\,\frac{{30}}{2},\,\frac{{40}}{2} \cr & \Rightarrow 5,\,10,\,15,\,20 \cr} $$