Question

A geometric sequence is given by the recursive rule $a_1=\frac{2}{3},\ a_n=3a_{n-1}$ . Give an explicit rule for the nth term of the sequence $b_1,\ b_2,\ b_3,\ \dots\ ,$ given that $b_1=a_1,\ b_2=a_3,\ b_3=a_5,\ \dots$ .

Answer 1

Answer:

$b_n = \frac{2}{27} (9^n)$

Step-by-step explanation:

Given

$a_1=\frac{2}{3},\ a_n=3a_{n-1}.$

Required

An explicit rule for $b_1,\ b_2,\ b_3,\ \dots$

Where $b_1=a_1,\ b_2=a_3,\ b_3=a_5,\ \dots$

We have:

$a_1=\frac{2}{3},\ a_n=3a_{n-1}.$

Calculate a2

$a_2 = 3a_{2-1}$

$a_2 = 3a_1$

$a_2 = 3 * \frac{2}{3}$

$a_2 = 2$

Calculate a3

$a_3 = 3a_{3-1}$

$a_3 = 3a_2$

$a_3 = 3 *2$

$a_3 = 6$

Calculate a4

$a_4 = 3a_{4-1}$

$a_4 = 3a_3$

$a_4 = 3*6$

$a_4 = 18$

Calculate a5

$a_5 = 3a_{5-1}$

$a_5 = 3a_4$

$a_5 = 3*18$

$a_5 = 54$

So:

$b_1=a_1,\ b_2=a_3,\ b_3=a_5,\ \dots$

$a_1=\frac{2}{3}$     $a_3 = 6$     $a_5 = 54$

The above sequence form a geometric sequence.

Calculate common ratio (r)

$r = \frac{b_3}{b_2}$

$r = \frac{a_5}{a_3}$

$r = \frac{54}{6}$

$r = 9$

So, the explicit formula is:

$b_n = b_1 * r^{n-1$

$b_1=a_1$, so:

$b_n = a_1 * r^{n-1$

$a_1=\frac{2}{3}$, so:

$b_n = \frac{2}{3} * 9^{n-1$

Split:

$b_n = \frac{2}{3} * \frac{9^n}{9}$

$b_n = \frac{2}{3*9} (9^n)$

$b_n = \frac{2}{27} (9^n)$

The explicit rule is: $b_n = \frac{2}{27} (9^n)$