+0  
 
0
364
2
avatar

The first two terms of a geometric progression are 27 and 9. How many terms are added up to get a sum of 364/9?

Guest Jan 24, 2015

Best Answer 

 #1
avatar+87301 
+10

We are trying to solve

27*∑ (1/3)^(n-1)   from 1 to n =  364/9

WolframAlpha shows that 6 terms are needed

Here are the sums from n = 1 to n = 6

27 + 9 + 3 + 1 + 1/3 + 1/9 = 364/9

 

CPhill  Jan 24, 2015
 #1
avatar+87301 
+10
Best Answer

We are trying to solve

27*∑ (1/3)^(n-1)   from 1 to n =  364/9

WolframAlpha shows that 6 terms are needed

Here are the sums from n = 1 to n = 6

27 + 9 + 3 + 1 + 1/3 + 1/9 = 364/9

 

CPhill  Jan 24, 2015
 #2
avatar+92781 
+5

a=27     r=1/3

27+9+3+1+(1/3)+(1/9)=364/9

 

 

$$\boxed{S_n=\frac{a(1-r^n)}{1-r}}\\\\\\
\frac{364}{9}=\frac{27(1-(1/3)^n)}{1-(1/3)}}\\\\
\frac{364}{9*27}=\frac{1-(1/3)^n}{2/3}}\\\\
\frac{364*2}{9*27*3}=1-(1/3)^n\\\\
\frac{728}{729}-1=-(1/3)^n\\\\
\frac{-1}{729}=-\frac{1}{3^n}\\\\
729=3^n\\\\
log_3(729)=log_3 3^n\\\\
n=log_3(729)\\\\
n=log729/log3\\\\
n=6$$

 

Thank you Chris for helping me spot my previous error   

Melody  Jan 25, 2015

8 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.