Question

find the common ratio, the 8th term, and the explicit formula. (a_n=a_1r^{n - 1}) 35) -1, 5, -25, 125, ... 36) -3, -9, -27, -81, ... 37) 4, -12, 36, -108, ... 38) -2, 10, -50, 250, ...

Explanation:

Step1: Find common ratio r

$r=\frac{a_{n + 1}}{a_{n}}=\frac{5}{-1}=- 5$

Step2: Find 8th term $a_8$

$a_n=a_1r^{n - 1}$, where $a_1=-1$, $r = - 5$, $n = 8$. So $a_8=(-1)\times(-5)^{8 - 1}=(-1)\times(-78125)=78125$

Step3: Find explicit formula

$a_n=a_1r^{n - 1}=(-1)\times(-5)^{n - 1}$

Step1: Find common ratio r

$r=\frac{a_{n + 1}}{a_{n}}=\frac{5}{-1}=- 5$

Step2: Find 8th term $a_8$

$a_n=a_1r^{n - 1}$, where $a_1=-1$, $r = - 5$, $n = 8$. So $a_8=(-1)\times(-5)^{8 - 1}=(-1)\times(-78125)=78125$

Step3: Find explicit formula

$a_n=a_1r^{n - 1}=(-1)\times(-5)^{n - 1}$

Step1: Find common ratio r

$r=\frac{a_{n+1}}{a_{n}}=\frac{-9}{-3}=3$

Step2: Find 8th term $a_8$

$a_n=a_1r^{n - 1}$, where $a_1=-3$, $r = 3$, $n = 8$. So $a_8=(-3)\times3^{8 - 1}=(-3)\times2187=-6561$

Step3: Find explicit formula

$a_n=a_1r^{n - 1}=(-3)\times3^{n - 1}=-3^{n}$

Step1: Find common ratio r

$r=\frac{a_{n + 1}}{a_{n}}=\frac{-12}{4}=-3$

Step2: Find 8th term $a_8$

$a_n=a_1r^{n - 1}$, where $a_1 = 4$, $r=-3$, $n = 8$. So $a_8=4\times(-3)^{8 - 1}=4\times(-2187)=-8748$

Step3: Find explicit formula

$a_n=a_1r^{n - 1}=4\times(-3)^{n - 1}$

Answer:

Common ratio: -5
8th term: 78125
Explicit formula: $a_n=(-1)\times(-5)^{n - 1}$

36) -3, -9, -27, -81, ...