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

#### Solution

Here, we are given that four number are in A.P., such that there 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 + 2d.

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

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

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

4a + 2d = 50

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

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

a + 2d = 4(a - d)

a + 2d = 4a - 4d

4d + 2d = 4a - a

6d = 3a

`d = 3/6 a` .....(2)

Now using (2) in (1) we get

`2a + 3/6 a = 25`

`(12a + 3a)/6 = 25`

15a = 150

`a = 150/15`

a = 10

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

`d = 3/6 (10)`

`d = 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 + 2d = 10 + (2)(5)

= 10 + 10

= 20

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