Question

Find the 6th term of the geometric sequence whose common ratio is 3/2 and whose first term is 6.

Answer 1

Answer:

The 6th term will be:

$a_6=\frac{729}{16}$

Step-by-step explanation:

Given

• a₁ = 6

• common ratio r = 3/2

To determine

a₆ = ?

A geometric sequence has a constant ratio r and is defined by

$a_n=a_1\cdot r^{n-1}$

substituting a₁ = 6, r = 3/2

$a_n=6\cdot \left(\frac{3}{2}\right)^{n-1}$

Determining 6th term

substituting n = 6 in the given equation

$a_n=6\cdot \left(\frac{3}{2}\right)^{n-1}$

$a_6=6\cdot \left(\frac{3}{2}\right)^{6-1}$

    $=6\cdot \frac{3^5}{2^5}$

    $=\frac{3^5\cdot \:6}{2^5}$

    $=\frac{3^5\cdot \:2\cdot \:3}{2^5}$

Cancel the common term

     $=\frac{3^6}{2^4}$

     $=\frac{729}{16}$

Therefore, the 6th term will be:

$a_6=\frac{729}{16}$