Question

four nos are in ap whose sum is 50 and the greatest no. is 4 times the least​

Answer 1

Answer:

Let the 4 numbers be a-2d, a-d, a, a+d, where a-2d is the least number and a+d is the highest number.

Greatest number is 4 times the least can be written as : a+d=4(a-2d)

a+d=4a-8d

4a-a=8d+d

3a=9d

a=9d/3

a=3d

The equation for the sum of numbers which are in AP can be written as: a-2d+a-d+a+a+d=50, -d and +d get canceled.

Now, the equation becomes a-2d+a+a+a=50.

Then a=3d should be substituted in the above equation.

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

3d-2d+3d+3d+3d=50

d+9d=50

10d=50d

d=50/10

d=5

substitute d=5 in the equation a=5d.

a=5×5=25.

Therefore the four numbers a-2d,a-d,a,a+d will be 25-(2×5),25-5,25,25+5. After simplifying it, the four numbers will be 15,20,25,30.