Question

Find four numbers in A.P., whose sum is 160 and in which the greatest number is 7 times the least.

Answer 1

Answer:

10, 30, 50, 70

Step-by-step explanation:

Let the first of the numbers be x and let the common difference be d>0.

Then the four numbers are x, x+d, x+2d, x+3d.

The sum is 160  ⇒  4x + 6d = 160  ⇒  2x + 3d = 80  ...(1)

The greatest is 7 times the least  ⇒  x+3d = 7x  ⇒  6x = 3d  ⇒  2x = d  ...(2)

Substituting (2) into (1) gives

  d + 3d = 80  ⇒  4d = 80  ⇒  d = 20

and then from (2)

  x = d/2 = 20/2 = 10.

So the four numbers are

  10, 30, 50, 70

Hope this helps!