Question

Write the rule for the nth term of the following sequence.

1 over 4 comma 1 over 2 comma 1 comma 2 comma 4 comma ...

Answer 1

Answer:

$2^{n-3}$ is the nth term of the given sequence.

Step-by-step explanation:

The sequence given is

$\frac{1}{4},\frac{1}{2},1,2,4 .....$

We can clearly see that 1st term is $\frac{1}{4}$ and 2nd term is $\frac{1}{2}$

2nd term is obtained by multiplying the 1st term by 2.

2nd term is $\frac{1}{2}$ and 3rd term is 1.

3rd term is obtained by multiplying the 2nd term by 2.

3rd term is 1 and 4th term is 2.

4th term is obtained by multiplying the 3rd term by 2.

Clearly, the given series is a Geometric Progression(GP) with

First term, $a = \frac{1}{4}$

Common Ratio, $r = 2$

We know that $n^{th}$ term for a GP is:

$a_n = ar^{n-1}$

Putting values of a and r

$a_n = \dfrac{1}{4}2^{n-1}\\a_n = \dfrac{1}{2^{2} }2^{n-1}\\a_n = 2^{n-1-2}\\a_n = 2^{n-3}$

Hence, $2^{n-3}$ is the nth term of the given sequence.