AAPL wrote:Veritas Prep
The first and second numbers in a sequence of numbers are plotted as the x and y coordinates, respectively, of a point on the coordinate plane, as are the third and fourth numbers, and all subsequent pairs in the sequence, and a line is formed by connecting these points, what would be the slope of the line?
1) Except for the first number, the sequence of numbers is formed by doubling the previous number and then subtracting 1.
2) The first number in the sequence is 3.
$$\left( {{a_1},{a_2}} \right)\,\,,\,\,\left( {{a_3},{a_4}} \right)\,\,,\,\,\left( {{a_5},{a_6}} \right)\,\,,\,\,\, \ldots \,\,\,\, \to \,\,\,{\text{all}}\,\,{\text{points}}\,\,{\text{belong}}\,\,{\text{to}}\,\,{\text{a}}\,\,{\text{certain}}\,\,{\text{line}}\,\,{\text{L}}$$
$${\text{?}}\,\,\,{\text{ = }}\,\,\,\,{\text{slop}}{{\text{e}}_{\,{\text{L}}}}\,\,\,\left( { = \,\,\frac{{{a_4} - {a_2}}}{{{a_3} - {a_1}}} = \,\frac{{{a_6} - {a_4}}}{{{a_5} - {a_3}}} = \ldots } \right)$$
$$\left( 1 \right)\,\,\,L\,\,\,:\,\,\,y = 2x - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 2\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{SUFF}}.$$
$$\left( 2 \right)\,\,\,{a_1}\, = 3\,\,\,\,\, \Rightarrow \,\,\,{\text{trivial}}\,\,{\text{geometric}}\,\,{\text{bifurcation}}\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.$$
Alternate (algebraic) argument for statement (1):
$$\left( 1 \right)\,\,{a_n} = 2 \cdot {a_{n - 1}} - 1\,\,\,,\,\,{\text{for}}\,\,{\text{all}}\,\,n \geqslant 2\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,? = \frac{{{a_4} - {a_2}}}{{{a_3} - {a_1}}} = \frac{{6\left( {{a_1} - 1} \right)}}{{\,3\left( {{a_1} - 1} \right)}}\, = 2\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\,$$
$$\left( * \right)\,\,\,\left\{ \matrix{
{a_4} = 2 \cdot {a_3} - 1 = 2\left( {2 \cdot {a_2} - 1} \right) - 1 = 4 \cdot {a_2} - 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{a_4} - {a_2}\,\, = \,\,\,3\left( {{a_2} - 1} \right) = 3\left[ {\left( {2 \cdot {a_1} - 1} \right) - 1} \right] = 6\left( {{a_1} - 1} \right) \hfill \cr
{a_3} = \,2 \cdot {a_2} - 1 = 2\left( {2 \cdot {a_1} - 1} \right) - 1 = 4 \cdot {a_1} - 3\,\,\,\, \Rightarrow \,\,\,\,\,\,{a_3} - {a_1}\,\, = \,\,\,3\left( {{a_1} - 1} \right) \hfill \cr} \right.$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
