image Geometric Sequence and Recursive Definition

Hello, SarahGr!

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1) In a Geometric Sequence: . $t_3 = 4\:\text{ and }\:t_6= \frac{4}{27}$ . . Find $t_{10}$

We have: . $\begin{array}{cccc}<br /> t_3 = 4 & \Rightarrow & t_{_1}r^2 \:=\:4 & [1] \\<br /> t_6 = \frac{4}{27} & \Rightarrow & t_{_1}r^5 \:=\:\frac{4}{27} & [2] \end{array}$

Divide [2] by [1]: . $\frac{t_{_1}r^5}{t_{_1}r^2} \:=\:\frac{\frac{4}{27}}{4} \quad\Rightarrow\quad r^3 \:=\:\frac{1}{27} \quad\Rightarrow\quad r \:=\:\frac{1}{3}$

Substitute into [1]: . $t_{_1}\left(\tfrac{1}{3}\right)^2 \:=\:4 \quad\Rightarrow\quad t_{_1} \:=\:36$

Therefore: . $t_{10} \;=\;t_{_1}r^9 \;=\;36\left(\tfrac{1}{3}\right)^9 \;=\;\frac{36}{19,\!683} \;=\;\frac{4}{2187}$

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2) Give a recursive definition for the sequence: . $1, 4, 13, 40, \hdots$

Take the differences of consecutive terms . . .

. . $\begin{array}{ccccccccc}\text{Sequence} & 1 && 4 && 13 && 40 \\<br /> \text{Difference} & & 3 & & 9 & & 27 \end{array}$

Each term is the preceding term plus a power of 3.

. . . . . $a_n \;=\;a_{n-1} + 3^{n-1}$