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0
413
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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
+4221
+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
+974
+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
+22172
+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:

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

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

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

Aug 28, 2018
edited by heureka  Aug 28, 2018
#4
+100516
+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

Aug 28, 2018