+0

# Arithmetic series

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141
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Let a1, a2, a3, ... be an arithmetic sequence. If a2/a4 = 3, what is a5/a3?

Jun 15, 2021

#1
-1

a5/a3 = 9.

Jun 15, 2021
#2
+2266
+1

a1 = x

a2 = x + r

a3 = x + 2r

a4 = x + 3r

a5 = x + 4r

(x + r)/(x + 3r) = a2/a3

I don't think it's possible to figured out (x + 4r)/(x + 2r).

=^._.^=

Jun 15, 2021
#3
+121063
+3

a2 / a 4  =   3

So

a4/ a2  =   1/3

a4  =  a2 /  3

a4  =   a2  +  2d  =   a2  / 3

So

a2  +  2d  =  a2  / 3

2d   =   -(2/3)a2

d  =   -(1/3)a2

a3   =    a2  +  d   =   a2  -  (1/3)a2 )   =    (2/3)a2

a5   =  a2   +   3d   =   a2  +  3( -(1/3)a2)   =    a2  -  a2  =  0

So

a5 / a3     =     0

Jun 15, 2021
#4
+26222
+4

Let a1, a2, a3, ... be an arithmetic sequence.
If a2/a4 = 3, what is a5/a3?

$$\begin{array}{|rcll|} \hline \dfrac{a_2}{a_4} &=& 3 \\ a_2 &=& 3a_4 \\ a_1+d &=& 3(a_1+3d) \\ a_1 + d &=& 3a_1+9d \\ -2a_1 &=& 8d \\ \mathbf{a_1} &=& \mathbf{-4d} \\ \hline \dfrac{a_5}{a_3} &=& x \\ a_5 &=& x*a_3 \\ a_1+4d &=& x*a_3 \quad | \quad \mathbf{a_1=-4d} \\ -4d+4d &=& x*a_3 \\ 0 &=& x*a_3 \\ x &=& \dfrac{0}{a_3} \\ \mathbf{x} &=& \mathbf{0} \\ \hline \end{array}$$

$$\dfrac{a_5}{a_3} = 0$$

Jun 15, 2021
#5
+121063
+1

I always feel good  when my answer  matches up with heureka's   !!!!!!

Jun 15, 2021