A question for CPhill

Am I allowed to answer??

Let \( a_1,a_2,a_3,\dots \)   be an arithmetic sequence. If  \(\dfrac{a_4}{a_2} = 3\), what is  \(\dfrac{a_5}{a_3}\) ?

\(d=(a_2-a_1)\\ so\\ a_1=a_1\\\ a_2=a_1+ (a_2-a_1)\\ a_3=a_1+ 2(a_2-a_1)\\ a_4=a_1+ 3(a_2-a_1)\\ a_5=a_1+ 4(a_2-a_1)\\~\\ \)

\(\dfrac{a_1+ 3(a_2-a_1)}{a_1+ (a_2-a_1)}=3\qquad find \qquad \dfrac{a_1+ 4(a_2-a_1)}{a_1+ 2(a_2-a_1)}\\ \dfrac{a_1+ 3(a_2-a_1)}{a_1+ (a_2-a_1)}=3\\ a_1+ 3(a_2-a_1)=3(a_1+ (a_2-a_1))\\ a_1+ 3(a_2-a_1)=3a_1+ 3(a_2-a_1)\\ a_1=3a_1\\ a_1=0\\ a_2=d\)

so

\(\dfrac{a_5}{a_3}=\dfrac{a+4d}{a+2d}=\dfrac{4a_2}{2a_2}=2\)

Let 

\(\mathbf{ a_1,a_2,a_3,\dots }\)

be an arithmetic sequence.

If \(\displaystyle \frac{a_4}{a_2} = 3,\)

 what is \(\displaystyle \frac{a_5}{a_3}\) ?

\(\begin{array}{|rcll|} \hline a_1 &=& a \\ a_2 &=& a+d \\ a_3 &=& a+2d \\ a_4 &=& a+3d \\ a_5 &=& a+4d \\ \hline \\ \dfrac{a_4}{a_2} &=& 3 \quad & | \quad a_4 = a+3d \qquad a_2 = a+d \\\\ \dfrac{a+3d}{a+d} &=& 3 \\\\ a+3d &=& 3(a+d) \\ a+\not{3d} &=& 3a+\not{3d} \\ a &=& 3a \quad & | \quad -a \\ 0 &=& 2a \quad & | \quad :2 \\ 0 &=& a \\ \mathbf{a} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|rclcr|} \hline a = 0: \\ \hline a_1 &=& 0 \\ a_2 &=& 0+d &=& d \\ a_3 &=& 0+2d &=& 2d \\ a_4 &=& 0+3d &=& 3d \\ a_5 &=& 0+4d &=& 4d \\ \hline \\ \dfrac{a_5}{a_3} &=& \dfrac{4\cdot \not{d}}{2\cdot \not{d}} \\\\ \dfrac{a_5}{a_3} &=& \dfrac{4}{2} \\\\ \mathbf{\dfrac{a_5}{a_3}} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

a4 / a2  =  3

Find

a5 / a3

a4 / a2   = 3   ⇒    a4  =  3a2   (1)

a2  +  2d   = a4      sub (1)  into the right side

a2  + 2d  =  3a2

2d  =   2a2

d  =  a2

So

a5  =  a2 + 3d   =   a2 + 3a2  =  4a2

a3  =  a2 + d    =   a2  + a2  =   2a2

So

a5 / a3  =   4a2 / 2a2   =  2

Probably just a coincidence......

All of you guys are correct! 

Sorry, Melody, I was about to type CPhill, Melody, and other users.

Yay! I'm back! 

Let \(a\)  be the first term, and let \(d\)  be the common difference. Then \(a_n = a + (n - 1)d\) for all \(n\) . In particular, \(a_4 = a + 3d\)  and \(a_2 = a + d\) ,so \(\frac{a + 3d}{a + d} = 3.\) Multiplying both sides by \(a+d\) , we get \(a + 3d = 3a + 3d\),   so \(a=0\)  .

Then, \(\frac{a_5}{a_3} = \frac{a + 4d}{a + 2d} = \frac{4d}{2d} = \boxed{2}.\)