Sequences: Plugging In
For each positive integer $$n$$, $$a_n$$ is defined such that $$a_{n+2} = 3a_n-a_{n+1}$$ and $$a_1 = 2$$.
In the table, select values for $$a_2$$ and $$a_4$$ that are together consistent with these conditions. Make only two selections, one in each column.
Incorrect.
If $$a_2 = -10$$, then
>$$a_3 = 3(2)-(-10) = 6+10 = 16$$
>$$a_4 = 3(-10)-16 = -30-16 = -46$$.
Since -46 is not one of the options, $$a_2$$ cannot be -10.
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Incorrect.
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Since $$a_3 = 6-a_2$$, you know that
>$$a_4 = 3(a_2)-(6-a_2)$$
Simplify this expression to find the relationship between $$a_2$$ and $$a_4$$.
Incorrect.
If $$a_2 = 3$$, then
>$$a_3 = 3(2)-(3) = 6-3= 3$$
>$$a_4 = 3(3)-3 = 9-3 = 6$$.
Since 6 is not one of the options, $$a_2$$ cannot be 3.
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Incorrect.
If $$a_2 = 5$$, then
>$$a_3 = 3(2)-(5) = 6-5 = 1$$
>$$a_4 = 3(5)-1= 15-1 = 14$$.
Since 14 is not one of the options, $$a_2$$ cannot be 5.
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Incorrect.
If $$a_2 = 12$$, then
>$$a_3 = 3(2)-(12) = 6-12 = -6$$
>$$a_4 = 3(12)-(-6) = 36+6 = 42$$.
Since 42 is not one of the options, $$a_2$$ cannot be 12.
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That's right!
Instead of trying all the possible pairs, you can find the relationship between $$a_2$$ and $$a_4$$.
Start with the fact that $$a_3 = 3(2) -a_2$$ and then use that to find an expression for $$a_4$$.
>$$a_3 = 6-a_2, \mbox{ thus } a_4 = 3(a_2)-(6-a_2)$$
>$$a_4 = 4a_2 - 6$$
If you try each of the answer choices for $$a_2$$, you can see which gives another one of the answer choices.
>$$a_4 = 4(-10) - 6 = -46$$
>$$a_4 = 4(-1) - 6=-10$$
>$$a_4 = 4(3) - 6 = 6$$
>$$a_4 = 4(5) - 6 = 14$$
>$$a_4 = 4(12) - 6 = 42$$
When $$a_2 = -1$$, then $$a_4 = -10$$. Since these are both answer choices, they must be the correct answers.
That's right!
-10
-1
3
5
12
-10
-1
3
5
12
