**Find a recurrence relation, with initial condition, that uniquely determines each of the following geometric progressions. a) 2, 10, 50, 250, . . . b) 7, 14/5, 28/25, 56/125, . . .**

The **Answer to the Question**

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**Here's the Solution to this Question**

Given Geometric progression;

a) 2,10,50,250,

Rewrite as,

$\begin{aligned} &a_{0}=2 \\ &a_{1}=10 \\ &a_{2}=50 \\ &a_{3}=250 \end{aligned}$

Now change each term of sequence in terms of a_{0};

$\begin{aligned} \text { since } a_{1} &=10=5 \times 2 \\ & \Rightarrow a_{1}=5 a_{0} \\ a_{2}=50 &=25 \times 2=25 a_{0} \\ & \Rightarrow a_{2}=(5)^{2} a_{0} \\ a_{3}=250 &=125 \times 2 \\ &=(5)^{3} a_{0} \end{aligned}$

Therefore, the recurrence relation is given by

$a_{n}=(5)^{n} a_{0} \quad \forall n \geqslant 1$

with Initial Condition $a_{0}=2$

b)

$\begin{array}{ll}7, 14 / 5,28 / 25,56 / 125, \ldots \\ a_{0}=7, a_{1}=14/5, a_{2}=28/25, a_{3}=56/125\end{array}$

Now,

$\begin{aligned} a_{1}=\frac{14}{5} &=\frac{2 \times 7}{5} \\ \Rightarrow a_{1} &=\frac{2}{5} a_{0} \end{aligned}$

$\begin{aligned} a_{2}=\frac{28}{25} &=\frac{4}{25} \times 7=\left(\frac{2}{5}\right)^{2} \times 7 \\ & \Rightarrow a_{2}=\left(\frac{2}{5}\right)^{2} a_{0} \mid \end{aligned}$

$\begin{aligned} a_{3}=\frac{56}{125} &=\frac{8 \times 7}{125}=\left(\frac{2}{5}\right)^{3} \times 7 \\ & \Rightarrow a_{3}=\left(\frac{2}{5}\right)^{3} a_{0} \end{aligned}$

Therefore Recurrence relation is given by;

$\ a_{n}=\underbrace{\left(\frac{2}{5}\right)^{n} a_{0} \quad \forall n \geqslant 1}_{\text {with } a_{0}=7}$

OR

$a_{n}=\frac{2}{5} a_{n-1}$