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December 20th, 2007, 03:18 AM  #1 
Newbie Joined: Nov 2007 Posts: 6 Thanks: 0  find the sum of gemetric sequence
∑ (5i+10) , i=1 bottm, 30 top (sorry, don't know how to make it look nicer) how do i find the sum of this? what is the equation and how do i find r? 
December 20th, 2007, 04:42 AM  #2 
Senior Member Joined: Oct 2007 From: France Posts: 121 Thanks: 1 
The sequence is arithmetic, not geometric. S=(10+5*1)+(10+5*2)+...+(10+5*30) ( there are 30 terms) =(10+10+..+10)+5*(1+2+..+30) =30*10+5*(1+2+..+30) You certainly knows that: 1+2+..+n=n(n+1)/2. From this, 1+2+..+30=30*31/2=465 and S=300+5*465=2 625. 
December 20th, 2007, 04:42 AM  #3 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
And r is? You could use the fact that the sum operator is linear and that: a ∑i= a(a+1)/2 i=1 a ∑c= a c i=1 
December 21st, 2007, 12:17 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,045 Thanks: 1618 
The sum of an arithmetic progression equals the product of the number of terms and the arithmetic mean of the first and last terms. For the problem given, that's 30 × (15 + 160)/2, i.e., 2625. 

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