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A.P.+ G.P.

:dontknow::lol::lolz::question::thumbsup::zomg: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.


Answer:5, 12, 19, 26

How to solve it? Show the steps.:confused::bban::bban:
Reply 1
Avatar for Krollo
Krollo
17
This looks like quite a fun question, I'll have a pop at it and get back to you.

Posted from TSR Mobile
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.


Answer:5, 12, 19, 26


As with the other question - can you show what you have done so far
Reply 3
Avatar for cucumberpj
cucumberpj
OP
Original post by Krollo
This looks like quite a fun question, I'll have a pop at it and get back to you.

Posted from TSR Mobile



Thanks! I urgently need it.
Reply 4
Avatar for cucumberpj
cucumberpj
OP
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.
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?
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...


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

Therefore

\dfrac{a+d-6}{a-2}=\dfrac{a+3d-7}{a+2d-2}
Reply 7
Avatar for cucumberpj
cucumberpj
OP
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?
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
Reply 9
Avatar for cucumberpj
cucumberpj
OP
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?
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

a+3d−2a+2d−7=a+2d−7a+d−6\dfrac{a+3d-2}{a+2d-7} = \dfrac{a+2d-7}{a+d-6}

and

a+2d−7a+d−6=a+d−6a−2\dfrac{a+2d-7}{a+d-6} = \dfrac{a+d-6}{a-2}

Two unknowns and two equations
Reply 11
Avatar for cucumberpj
cucumberpj
OP
Original post by TenOfThem
Well, you can do it twice

a+3d−2a+2d−7=a+2d−7a+d−6\dfrac{a+3d-2}{a+2d-7} = \dfrac{a+2d-7}{a+d-6}

and

a+2d−7a+d−6=a+d−6a−2\dfrac{a+2d-7}{a+d-6} = \dfrac{a+d-6}{a-2}

Two unknowns and two equations


thanks. I solved it. This method is easier to understand.:colondollar:

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