Question

Find the 12th term of the geometric sequence 5, -25, 125, ...5,−25,125,...

Answer 1

Answer:

• $a_{12}=-244140625$

Step-by-step explanation:

Considering the geometric sequence

$5,-25,\:125,\:...$

$a_1=5$

As the common ratio '$r$' between consecutive terms is constant.

$\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}$

$r=\frac{-25}{5}=-5$

$r=\frac{125}{-25}=-5$

The general term of a geometric sequence is given by the formula:  

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

where $a_1$ is the initial term and $r$ the common ratio.

Putting $n = 12$ , $r = -5$ and $a_1=5$ in the general term of a geometric sequence to determine the 12th term of the sequence.

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

$a_n=5\left(-5\right)^{n-1}$

$a_{12}=5\left(-5\right)^{12-1}$

      $=5\left(-5^{11}\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

       $=-5\cdot \:5^{11}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

        $=-5^{1+11}$     ∵ $5\cdot \:5^{11}=\:5^{1+11}$

        $=-244140625$

Therefore,

• $a_{12}=-244140625$