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

Sagot :

Absor201idn

Answer:

The 6th term will be:

[tex]a_6=\frac{729}{16}[/tex]

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

[tex]a_n=a_1\cdot r^{n-1}[/tex]

substituting a₁ = 6, r = 3/2

[tex]a_n=6\cdot \left(\frac{3}{2}\right)^{n-1}[/tex]

Determining 6th term

substituting n = 6 in the given equation

[tex]a_n=6\cdot \left(\frac{3}{2}\right)^{n-1}[/tex]

[tex]a_6=6\cdot \left(\frac{3}{2}\right)^{6-1}[/tex]

    [tex]=6\cdot \frac{3^5}{2^5}[/tex]

    [tex]=\frac{3^5\cdot \:6}{2^5}[/tex]

    [tex]=\frac{3^5\cdot \:2\cdot \:3}{2^5}[/tex]

Cancel the common term

     [tex]=\frac{3^6}{2^4}[/tex]

     [tex]=\frac{729}{16}[/tex]

Therefore, the 6th term will be:

[tex]a_6=\frac{729}{16}[/tex]

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