cucumberpj
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The first three of four integers are in an A.P. and tha last three are in a G.P..Find these four numbers, given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.


please kindly show the steps!thanks
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TenOfThem
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(Original post by cucumberpj)
The first three of four integers are in an A.P. and tha last three are in a G.P..Find these four numbers, given that the sum of the first and the last integers is 37 and the sum of the two integers in the middle is 36.
That is a really nice question

Can you show what you have done so far
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cucumberpj
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(Original post by TenOfThem)
That is a really nice question

Can you show what you have done so far
A.P. :a, a+d, a+2d

a+d+a+2d=36
2a+3d=36

I don't know how to solve further.:confused:
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TenOfThem
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(Original post by cucumberpj)
A.P. :a, a+d, a+2d

a+d+a+2d=36
2a+3d=36

I don't know how to solve further.:confused:
ok - now I will say that this approach worked really well on the other question

I took a different approach with this one

I will start you off - see if you can do anything with this

If we let the numbers be a, b, 36-b, 37-a

Using common difference you know that 36-b - b = b - a

Then we can apply the constant ratio principle

so  \dfrac{37-a}{36-b} = \dfrac{36-b}{b}
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cucumberpj
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(Original post by TenOfThem)
ok - now I will say that this approach worked really well on the other question

I took a different approach with this one

I will start you off - see if you can do anything with this

If we let the numbers be a, b, 36-b, 37-a

Using common difference you know that 36-b - b = b - a

Then we can apply the constant ratio principle

so  \dfrac{37-a}{36-b} = \dfrac{36-b}{b}

Thanks! I have solved it. Really appreciate.
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TenOfThem
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(Original post by cucumberpj)
Thanks! I have solved it. Really appreciate.
np

it is important to remember that constant difference and constant ratio can be used in the way I showed
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