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April 29th, 2015, 02:17 PM   #1
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Very challenging sequence 2

Another challenging sequence problem. Please help me in the resolution!
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April 29th, 2015, 02:18 PM   #2
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Sequence challenging problem 3

This is the last challenging problem I will post today. I don't really know how to solve this. I will appreciate your help!
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Last edited by skipjack; April 29th, 2015 at 11:12 PM.
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April 29th, 2015, 02:40 PM   #3
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Not all that "challenging".
For the first, it is easy to see that the "a" sequence, $\displaystyle a_1= 7$, $\displaystyle d= 3$, is 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, ....

The "b" sequence, with "$\displaystyle b_4= 21$, $\displaystyle b_{10}= 51$ then we have $\displaystyle b_0+ 3d= 21$ and $\displaystyle b_0+ 9d= 51$. Subtracting the first equation from the second, the $\displaystyle b_0$ terms cancel leaving 6d= 30 so d= 5. Then $\displaystyle b_0+ 3d= b_0+ 15= 21$ so $\displaystyle b_0= 21- 15= 6$. That is, the sequence is 6, 11, 16, 21, 26, 31, 36, 41, 46, ....

It is easy to see that the numbers in both sequences are 16, 31, 46, etc. 31- 16= 15 and 46- 31= 15= 3(5).


The second one, 8, x, y are three consecutive terms in an arithmetic sequence then x- 8= y- x so that y= 2x- 8. If, in addition, y, $\displaystyle \sqrt{42}$, x are three consecutive terms in a geometric sequence, then $\displaystyle \frac{y}{\sqrt{42}}= \frac{\sqrt{42}}{x}$ so that xy= 42.

Solve the two equations y= 2x- 8 and xy= 42.
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Last edited by Country Boy; April 29th, 2015 at 02:47 PM.
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April 29th, 2015, 03:45 PM   #4
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Quote:
Originally Posted by Country Boy View Post
Not all that "challenging".
For the first, it is easy to see that the "a" sequence, $\displaystyle a_1= 7$, $\displaystyle d= 3$, is 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, ....

The "b" sequence, with "$\displaystyle b_4= 21$, $\displaystyle b_{10}= 51$ then we have $\displaystyle b_0+ 3d= 21$ and $\displaystyle b_0+ 9d= 51$. Subtracting the first equation from the second, the $\displaystyle b_0$ terms cancel leaving 6d= 30 so d= 5. Then $\displaystyle b_0+ 3d= b_0+ 15= 21$ so $\displaystyle b_0= 21- 15= 6$. That is, the sequence is 6, 11, 16, 21, 26, 31, 36, 41, 46, ....

It is easy to see that the numbers in both sequences are 16, 31, 46, etc. 31- 16= 15 and 46- 31= 15= 3(5).


The second one, 8, x, y are three consecutive terms in an arithmetic sequence then x- 8= y- x so that y= 2x- 8. If, in addition, y, $\displaystyle \sqrt{42}$, x are three consecutive terms in a geometric sequence, then $\displaystyle \frac{y}{\sqrt{42}}= \frac{\sqrt{42}}{x}$ so that xy= 42.

Solve the two equations y= 2x- 8 and xy= 42.


Thanx a lot for the detailed explanation
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