Question

A sequence can be generated by using an = 4a(n - 1)) where a= 6 and n is a whole

Answer 1

Given:

The sequence can be generated by

$a_n=4a_{(n - 1)}$

Where $a_1= 6$ and $n$ is a whole number greater than 1.

To find:

The first four terms of the given sequence.

Solution:

We have,

$a_n=4a_{(n - 1)}$          ...(i)

It is given that $a_1= 6$. So, for $n=2$, we get

$a_2=4a_{(2 - 1)}$

$a_2=4a_1$

$a_2=4(6)$

$a_2=24$

Putting $n=3$ in (i), we get

$a_3=4a_{(3- 1)}$

$a_3=4a_2$

$a_3=4(24)$

$a_3=96$

Putting $n=4$ in (i), we get

$a_4=4a_{(4- 1)}$

$a_4=4a_3$

$a_4=4(96)$

$a_4=384$

The first four terms of the given sequence are 6, 24, 96,384.

Therefore, the correct option is C.