4 numbers are in an AP and their sum is 20. What is the greatest number if the greatest is 4 times least?

Let the smallest number be x and let the common difference be d

The numbers are

x,x+d.x+2d and x+3d

But it is given that x+3d=4x

or,-3x=-3d

or x=d

Hence the numbers are x,2x,3x and 4x

By the condition of the problem

x+2x+3x+4x=20

or 10x=20

x=2

hence the greatest number is 4*2=8

2 + 4 + 6 + 8 = 20

The greatest number is 8

2 + 4 + 6 + 8 = 20

The arithmetic progression (AP) must look like x,x+y,x+2*y,x+3*y for some integers x and y. From your conditions we get

x+(x+y)+(x+2*y)+(x+3*y)=20

and

x+3*y=4*x

Solving the second equation, we get x=y. Plugging this into the first equation, we get

x+2*x+3*x+4*x=20,

which gives us x=2, thus y=2.

So the greatest number is

x+3*y=2+3*2

=8

dunno what an AP is but the greatest number is 8, either 2, 4, 6, 8 or 2, 3, 7, 8 unless you can use the same number more than once, then the sequence could be different, but 8 would still be the answer you are looking for. I hope your maths teacher doesnt come on yahoo answers. or your mom, then she would know you are gay

8

2, 4, 6, 8

What is AP?

2,4,6,8...

i worked this out through trial and error starting with 4,8,12,16... so i can't really explain much but i worked out that the numbers had to be quite small and so did the common differences (and also that the cd's had to sort of relate to 4...if that helps in any way) - you did mean 'AP' as in 'arithmetic progression' in sequences?

AP?

The answer must be 8

2,4,6,8

8,6,4,2. x