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**Soroban** Hello, mrszone!

It is not a geometric series . . . .$\displaystyle x_1=5,\;x_2 = \text{-}8,\;x_3 =\text{-}23.6 \;\hdots$

. . $\displaystyle \begin{array}{cccccc}\text{We have:} & x_{n+1} &=& 1.2x_n - 14 & {\color{blue}[1]}\\$$\displaystyle

\text{and:} & x_{n+2} &=& 1.2x_{n+1} - 14 & {\color{blue}[2]} \end{array}$

Subtract [2] - [1]: .$\displaystyle x_{n+2} - x_{n-1} \;=\;1.2x_{n+1} - 1.2x_n \quad\Rightarrow\quad x_{n+2} - 2.2x_{n+1} + 1.2x_n \:=\:0$

Suppose the function is exponential: .$\displaystyle x_n \:=\:Y^n$

. . The equation becomes: .$\displaystyle Y^{n+2} - 2.2Y^{n+1} + 1.2Y^n \:=\:0$

. . which factors: .$\displaystyle Y^n\left(Y^2 - 2.2Y + 1.2\right) \:=\:0 \quad\Rightarrow\quad Y^n(Y - 1)(Y - 1.2) \;=\;0$

. . and has nonzero roots: .$\displaystyle Y \:=\:1,\:1.2$

$\displaystyle \text{The function, }f(n)\text{, is either: }\:1^n\,\text{ or }\,1.2^n$

Form a linear combination of the two functions: .$\displaystyle f(n) \;=\;A + B(1.2)^n$

From the first two values of the sequence:

. . $\displaystyle \begin{array}{cccccc}f(1)=5\!: & A + 1.2B &=& 5 & {\color{blue}[3]} \\$$\displaystyle

f(2)=\text{-}8\!: & A + 1.44B &=&\text{-}8 & {\color{blue}[4]} \end{array}$

Subtract [4] - [3]: .$\displaystyle 0.24B \:=\:\text{-}13 \quad \Rightarrow\quad\boxed{ B \:=\:-\frac{325}{6}}$

Substitute into [3]: .$\displaystyle A + (1.2)\left(-\tfrac{325}{6}\right) \:=\: 5 \quad\Rightarrow\quad\boxed{ A \:=\:70}$

Therefore: . $\displaystyle f(n) \;=\;70 -\frac{325}{6}(1.2)^m$