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# A.P.+ G.P. Watch

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1. Find four consecutive numbers in an A.P. such that when 2, 6, 7 and 2 are subtracted from these numbers respectively., the numbers are in geometric progression.

How to solve it? Show the steps.
2. This looks like quite a fun question, I'll have a pop at it and get back to you.

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3. (Original post by cucumberpj)
Find four consecutive numbers in an A.P. such that when 2, 6, 7 and 2 are subtracted from these numbers respectively., the numbers are in geometric progression.

As with the other question - can you show what you have done so far
4. (Original post by Krollo)
This looks like quite a fun question, I'll have a pop at it and get back to you.

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Thanks! I urgently need it.
5. (Original post by TenOfThem)
As with the other question - can you show what you have done so far
I really have no idea of it.
6. (Original post by cucumberpj)
I really have no idea of it.
Are you saying that you do not know what an AP or a GP is?
7. I haven't done it yet myself, but this is how I would start.

You know the numbers are in AP, so you can represent them as a, a+d...

Hint

Following on, from the rules of GP, the ratio between each consecutive term must be the same.

Therefore

8. Let x, x+d, x+2d, and x+3d be the numbers in the arithmetic sequence.
Let the numbers of the corresponding geometric sequence be:
a, ar, ar^2, and ar^3.

4 equation, 4 unknowns:
x = a + 2 [1]
x + d = ar + 6 [2]
x + 2d = ar^2 + 7 [3]
x + 3d = ar^3 + 2 [4]

Substitute [1] into [2] and solve for d:
a + 2 + d = ar + 6
d = a(r - 1) + 4

Substitute d and [1] into [3] and solve for a:
a + 2 + 2[a(r - 1) + 4] = ar^2 + 7
3 = ar^2 - 2ar + a
3 = a(r - 1)^2
3/(r - 1)^2 = a

Substitute a, d, and [1] into [4] and solve for r:
3/(r - 1)^2 + 2 + 3[(3/(r - 1)^2)(r - 1) + 4] =(3/(r - 1)^2)r^3 + 2
3 + 12(r - 1)^2 + 9(r - 1) = 3r^3
0 = 3(r^3 - 4r^2 + 5r - 2)
0 = (r - 2)(r - 1)^2
r = 1 or 2

Substituting r = 1 into the expression for a won't work since we would be dividing by zero, so r = 2 is the only solution.

a = 3/(2 - 1)^2 = 3

d = 3(2 - 1) + 4 = 7

x = 3 + 2 = 5

x + d = 12
x + 2d = 19
x + 3d = 26

any other simple method to solve it?
9. (Original post by cucumberpj)
Let x, x+d, x+2d, and x+3d be the numbers in the arithmetic sequence.
Use this and then the constant ratio idea that I showed you on the other thread

You can do that twice here and then you have a simple quadratic
10. (Original post by TenOfThem)
Use this and then the constant ratio idea that I showed you on the other thread

You can do that twice here and then you have a simple quadratic
x, x+d, x+2d, and x+3d
=a , a+d, a+2d, a+3d

G.P. : a-2 , a+d-6, a+2d-7, a+3d-2
then apply common ratio, the equation has two unknowns, how to solve it?
11. (Original post by cucumberpj)
x, x+d, x+2d, and x+3d
=a , a+d, a+2d, a+3d

G.P. : a-2 , a+d-6, a+2d-7, a+3d-2
then apply common ratio, the equation has two unknowns, how to solve it?
Well, you can do it twice

and

Two unknowns and two equations
12. (Original post by TenOfThem)
Well, you can do it twice

and

Two unknowns and two equations
thanks. I solved it. This method is easier to understand.

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