Question

find the four miners in an AP whose sum is 50 and in which the greatest number is 4 times the least​

Answer 1

Step-by-step explanation:

Let the four numbers be

a-3d,a-d,a+d,a+3d (imp step)

acc to qn

a-3d+a-d+a+d+a+3d=50

4a=50

a=25/2

Let d be positive, then the greatest number is a+3d

and

least number is a-3d

acc to qn

a+3d=4(a-3d)

15d=3a

5d=a

d=a/5=1/5*(25/2)=5/2

so the numbers in A.P. are

5,10,15,20

we see that their sum is 50

and

the greatest number i.e. 20 is four times the smallest number i.e. 5