How Do You Solve These Tricky Arithmetic Sequence Problems?

[asd1249jf]

Homework Statement

Problem 1:

Two Arithmetic Sequences are given.

[lat]a_n = 200,196,192,188,184...[/lat]
[ltaex]b_n = 100,103,106,109,112...[/itex]

For integers l,m, find the number of pairs consisting of (l,m) which satifies condition a_l = b_m




Problem 2:

Three terms, sin(x), sqrt(3)/4, cos(x) form an arithmetic sequence. Find 2 |tan(x) + 1/tan(x)|

Homework Equations

a_n = a + (n-1)d

The Attempt at a Solution

For problem 1:

The equations formed for both arithmetic sequences are

a_n = -4n + 204
b_n = 3n + 97

I thought about making them equal to find the exact n which both a_n and b_n is equal of but I'm pretty sure that's not the correct way to do it. What should I do?


And for Problem 2:

since three terms are in incremental sequence, we can find that

sqrt(3)/4 = (sin(x) + cos(x)) / 2

which yields

sqrt(3)/2 = sin(x) + cos(x)

I feel like I need to do something with trigonometric identity to substitute that, so I tried:

2|tan^2(x) + 1 / tan(x)|

= 2|sec^2(x) / tan(x)|

= 2|1/cos(x)(sin(x)|

which didn't get me anywhere.
Any suggestions?

 

[asd1249jf]

What.. is latex tag broken..?

 

[rl.bhat]

a_l = b_n
200 - ( l -1)*4 = 100 + (n-1)*3
Simplifying you will get
107 = 4l + 3n
Try n = 9, 19, 29,...so on.

 

[asd1249jf]

a_l = b_n
200 - ( l -1)*4 = 100 + (n-1)*3
Simplifying you will get
107 = 4l + 3n
Try n = 9, 19, 29,...so on.

I understand how you simplified the equation.. but where did you get the idea of plugging 9,19,29 for n?

nvm i see it now. TY.

 

[asd1249jf]

Uh so anyone got an idea about second problem?

 

[HallsofIvy]

Three terms, sin(x), sqrt(3)/4, cos(x) form an arithmetic sequence. Find 2 |tan(x) + 1/tan(x)|

So sin(x)+ d= sqrt(3)/4, and sin(x)+ 2d= cos(x). From sin(x)+ d= \sqrt(3)/4, d= sqrt(3)/4- sin(x). Put that into sin(x)+ 2d= cos(x): sin(x)+ 2sqrt(3)/4- 2sin(x)= cos(x)

cos(x)= 2sqrt(3)/4- sin(x) Since we also know that sin^2(x)+ cos^2(x)= 1, we can solve for sin(x) and cos(x).

 

[rl.bhat]

sin(x) + cos(x) = sqrt(3)/2

Square both the sides,

sin^2(x) + co^2(x) + 2*sin(x)cos(x) = 3/4

2*sin(x)cos(x) = (3/4) - 1 = -(1/4)

Now proceed.