ANOTHER probability problem...please help! :P

[1a2s3d4f5g6h7j8k9l]

Hey all...here's the problem:

You are given two arithmetic sequences:

1) 3, 14, 25, 36,...
2) 2, 9, 16, 23,...

Find the first 5 numbers that the two sequences have in common.

Can someone please teach me how to do this problem? Thanks!

 

[Unknown008]

The terms increase by a constant number. So, you need to add that certain number to the previous number to get the next number. Can you try it out with what I told you?

If not, post back

 

[soroban]

Hello, 1a2s3d4f5g6h7j8k9l!

You are given two arithmetic sequences:

. . \(\displaystyle 1)\;3,14,25,36,\hdots\)

. . \(\displaystyle 2)\;2,9,16,23,\hdots\)

Find the first 5 numbers that the two sequences have in common.

Sequence #1 has the general term: .\(\displaystyle a_m \:=\:11m-8\)
Sequence #2 has the general term: .\(\displaystyle b_n \:=\:7n - 5\)

We want to know when \(\displaystyle a_m = b_n.\)


We have: .\(\displaystyle 11m - 8 \:=\:7n-5 \quad\Rightarrow\quad n \:=\:\dfrac{11m-3}{7} \:=\:m + \dfrac{4m-3}{7}\)

Since \(\displaystyle n\) is an integer, \(\displaystyle 4m-3\) is divisible by 7.
. . Let: \(\displaystyle 4m-3 \:=\:7a\), for some integer \(\displaystyle a.\)

And we have: .\(\displaystyle m \:=\:\dfrac{7a+3}{4} \:=\:a + \dfrac{3a+3}{4} \:=\:a + \dfrac{3(a+1)}{4}\) .[1]

Since \(\displaystyle m\) is an integer, \(\displaystyle a+1\) is divisible by 4.
. . Hence: .\(\displaystyle a \:=\:3, 7, 11, 15, 19,\,\hdots\)

Substitute into [1]: .\(\displaystyle m \:=\:6, 13, 20, 27, 34,\,\hdots\)


Therefore: .\(\displaystyle \begin{Bmatrix}a_6 &=& 58 &=& b_9 \\ a_{13} &=& 135 &=& b_{20} \\ a_{20} &=& 212 &=& b_{31} \\ a_{27} &=& 289 &=& b_{42} \\ a_{34} &=& 366 &=& b_{53} \end{Bmatrix}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


There is a faster way.

Note that they are in-synch every \(\displaystyle 11\!\cdot\!7 \,=\,77\) "steps".

Determine the value the first time they are equal,
. . and add 77 repeatedly.

I'll let someone else explain it.

 

[1a2s3d4f5g6h7j8k9l]

Thanks!

Thank you Unknown008 and Soroban for replying to the thread(s)...it helped a lot!!!