Question

2. Find the sum of the first 6 terms of the geometric series.

Answer 1

Answer:

(a)$S_6=1092$

(b)$S_5=363$

Step-by-step explanation:

Question 1

Given the geometric series

3+15+75+375+...

$a_1=3, r=15/3=75/15=5\\ Therefore using:\\S_n=\dfrac{a_1(1-r^n)}{1-r} \\S_6=\dfrac{3(1-3^6)}{1-3}=\dfrac{3(-728)}{-2}=\dfrac{-2184}{-2}\\\\S_6=1092$

Question 2

Given the series: $\sum_{n=1}^5 3^n$

$3^1=3;3^2=9;3^3=27\\ The series is 3+9+27+\cdots\\ Therefore: r=9/3=27/9=3$

$a_1=3, r=3\\ Therefore using:\\S_n=\dfrac{a_1(1-r^n)}{1-r} \\S_5=\dfrac{3(1-3^5)}{1-3}=\dfrac{3(-242)}{-2}=\dfrac{-726}{-2}\\\\S_5=363$