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I have the following terms of an arithmetic sequence: $\frac{1}{2}, x-1, 3x, \ldots$. Solve for $x$.

 Aug 28, 2018
 #1
avatar+3994 
+1

Isn't the gap between two consecutive numbers, the same? So, \(3x-x-1=\frac{1}{2}, 2x-1=\frac{1}{2}, 2x=\frac{3}{2}, x=\frac{3}{4}\)

 Aug 28, 2018
 #2
avatar+972 
+1

Consecutive terms of an arithemtic sequence have a common difference. 

 

We can write the equaiton:

 

\((x-1)-\frac12=3x-(x-1) \\ x-1.5=2x+1\\ x=-2.5\)

 

Then, we can plug this value into the terms to check if they have a common difference. 

 

The sequence goes: \(\frac12, -2.5-1, 3\cdot(-2.5) \Rightarrow0.5, -3.5, -7.5\)

 

The common difference is 4, so the value works!

 

I hope this helped,

 

Gavin. 

 Aug 28, 2018
 #3
avatar+21848 
+2

I have the following terms of an arithmetic sequence:

\(\frac{1}{2}, x-1, 3x, \ldots\)

$\frac{1}{2}, x-1, 3x, \ldots$.

Solve for \(x\).

 

Formula of an arithmetic sequence:

\(\begin{array}{|rcll|} \hline \begin{vmatrix} a_i & a_j & a_k \\ i & j & k \\ 1 & 1 & 1 & \\ \end{vmatrix} = 0 \\\\ a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\ \hline \end{array} \)

 

\(\text{Set $i=1$, $\ j=2$, $\ k=3$ } \\ \text{Set $a_i=a_1=\frac12$, $\ a_j=a_2=x-1$, $\ a_k=a_3=3x$ } \)

 

\(\begin{array}{|rcll|} \hline a_i(j-k)+a_j(k-i)+a_k(i-j) &=& 0 \\\\ \frac12(2-3)+(x-1)(3-1)+3x(1-2) &=& 0 \\ \frac12(-1)+(x-1)(2)+3x(-1) &=& 0 \\ -\frac12+2x-2-3x &=& 0 \\ -\frac52-x &=& 0 \\ \frac52+x &=& 0 \\ x &=& -\frac52 \\ \mathbf{x} &\mathbf{=}& \mathbf{-2.5} \\ \hline \end{array} \)

 

laugh

 Aug 28, 2018
edited by heureka  Aug 28, 2018
 #4
avatar+98125 
+1

We have two  equations in two unknowns...call the common  difference, d

 

(1/2) +  d  = x -1  ⇒    (1/2) + 1  = x - d      (1)

x -1 + d  = 3x    ⇒       -1 =    2x  - d    ⇒  1 = -2x + d      (2)

 

Add (1)  and (2)  and we get

 

(5/2)  = - x

 

-5/2  =  x

 

And  d  =  (3/2)  = (-5/2) - d ⇒   d  =  -5/2  - 3/2  = -8/2   = -4

 

Proof

 

1/2 , -5/2  -1 , 3 (-5/2)

 

1/2 , -7/2 , -15/2         with   d  =  -4

 

 

cool cool cool

 Aug 28, 2018

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