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} a 0 = 2 a 1 = 10 a 2 = 50 a 3 = 250
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} since a 1 a 2 = 50 a 3 = 250 = 10 = 5 × 2 ⇒ a 1 = 5 a 0 = 25 × 2 = 25 a 0 ⇒ a 2 = ( 5 ) 2 a 0 = 125 × 2 = ( 5 ) 3 a 0
Therefore, the recurrence relation is given by
a_{n}=(5)^{n} a_{0} \quad \forall n \geqslant 1 a n = ( 5 ) n a 0 ∀ n ⩾ 1
with Initial Condition a_{0}=2 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} 7 , 14/5 , 28/25 , 56/125 , … a 0 = 7 , a 1 = 14/5 , a 2 = 28/25 , a 3 = 56/125
Now,
\begin{aligned} a_{1}=\frac{14}{5} &=\frac{2 \times 7}{5} \\ \Rightarrow a_{1} &=\frac{2}{5} a_{0} \end{aligned} a 1 = 5 14 ⇒ a 1 = 5 2 × 7 = 5 2 a 0
\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} a 2 = 25 28 = 25 4 × 7 = ( 5 2 ) 2 × 7 ⇒ a 2 = ( 5 2 ) 2 a 0 ∣
\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} a 3 = 125 56 = 125 8 × 7 = ( 5 2 ) 3 × 7 ⇒ a 3 = ( 5 2 ) 3 a 0
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} a n = with a 0 = 7 ( 5 2 ) n a 0 ∀ n ⩾ 1
OR
a_{n}=\frac{2}{5} a_{n-1} a n = 5 2 a n − 1