Question

determine if the sequence is geometric if it is find the common ratio the 8th term and the explicit formula can someone check my work

Answer 1

Answer:

You can check the answer by yourself after seeing the below answer.

Step-by-step explanation:

A sequence is geometric only if has common ratio i.e. $r=\frac{a2}{a1} =\frac{a3}{a2}$ whiere a1,a2 and a3 are first,second and third term of the sequence respectively.

1) Common ratio $r=\frac{-18}{-3}=\frac{-108}{-18}=6$

 Explicit formula $a_{n} =a_{1} .r^{n-1}$

Now using the above formula, we can find $8^{th} term=-3*(6)^{8-1} =-3*6^{7}=-839808.$

2) Here

$\frac{a2}{a1}\neq \frac{a3}{a2}\\\frac{-2}{-4}\neq \frac{1}{-2}\\\frac{1}{2}\neq\frac{1}{-2}$ so this is not geometric sequence so no need to proceed further.

3) Common ratio $r=\frac{15}{-3}=\frac{-75}{15}=-5$

 Explicit formula $a_{n} =a_{1} .(-5)^{n-1}$

Now using the above formula, we can find $8^{th} term=-3*(-5)^{8-1} =-3*(-5)^{7}=234375.$

4) Common ratio $r=\frac{-6}{1}=\frac{36}{-6}=-6$

 Explicit formula $a_{n} =a_{1} .(-6)^{n-1}$

Now using the above formula, we can find $8^{th} term=1*(-6)^{8-1} =1*(-6)^{7}=-279936.$

5)Common ratio $r=\frac{-8}{4}=\frac{16}{-8}=-2$

 Explicit formula $a_{n} =a_{1} .(-2)^{n-1}$

Now using the above formula, we can find $8^{th} term=4*(-2)^{8-1} =4*(-2)^{7}=-512.$

6)Common ratio $r=\frac{-8}{-2}=\frac{-32}{-8}=4$

 Explicit formula $a_{n} =a_{1} .(4)^{n-1}$

Now using the above formula, we can find $8^{th} term=-2*(4)^{8-1} =-2*(4)^{7}=-32768.$