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Directions: Drag each tile to the correct box.Put the recursive formulas below in order from least to greatest according to the value of their 10th terms.For all of the formulas, let n be equal to the whole numbers greater than or equal to one.

Directions Drag Each Tile To The Correct BoxPut The Recursive Formulas Below In Order From Least To Greatest According To The Value Of Their 10th TermsFor All O class=

Sagot :

Solving for the 10th term for each of the recursive sequence

First sequence

[tex]\begin{gathered} a_1=32 \\ a_{n+1}=-5+a_n \\ \\ \text{This can be converted to} \\ a_n=a_1+(n-1)(-5) \\ \\ \text{Substitute }n=10 \\ a_{10}=32+(10-1)(-5) \\ a_{10}=32+(9)(-5) \\ a_{10}=32-45 \\ a_{10}=-13 \end{gathered}[/tex]

Second sequence

[tex]\begin{gathered} a_1=2048 \\ a_{n+1}=-\frac{1}{2}a_n \\ \\ \text{This can be converted to} \\ a_n=a_1\cdot\Big(-\frac{1}{2}\Big)^{n-1} \\ \\ \text{Substitute }n=10 \\ a_{10}=2048\cdot\Big(-\frac{1}{2}\Big)^{10-1} \\ a_{10}=2048\cdot\Big(-\frac{1}{2}\Big)^9 \\ a_{10}=-4 \end{gathered}[/tex]

Third sequence

[tex]\begin{gathered} a_1=0.125 \\ a_{n+1}=2a_n \\ \\ \text{This can be converted to} \\ a_n=a_1\cdot2^{n-1} \\ \\ \text{Substitute }n=10 \\ a_{10}=0.125\cdot2^{10-1} \\ a_{10}=0.125\cdot2^9 \\ a_{10}=64 \end{gathered}[/tex]

Fourth sequence

[tex]\begin{gathered} a_1=-7\frac{2}{3} \\ a_{n+1}=a_n+1\frac{2}{3} \\ \\ \text{This can be converted to} \\ a_n=a_1+(n-1)\Big(1\frac{2}{3}\Big) \\ \\ \text{Substitute }n=10 \\ a_{10}=-7\frac{2}{3}+(10-1)\Big(1\frac{2}{3}\Big) \\ a_{10}=\frac{-23}{3}+(9)\Big(\frac{5}{3}\Big) \\ a_{10}=-\frac{23}{3}+\frac{45}{3} \\ a_{10}=\frac{22}{3} \\ a_{10}=7\frac{1}{3} \end{gathered}[/tex]

Arranging the formulas from least to greatest according to their 10th terms, we have the following:

First Sequence → Second Sequence → Fourth Sequence → Third Sequence