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September 20th, 2015, 06:43 AM  #1 
Newbie Joined: Sep 2015 From: United Kingdom Posts: 8 Thanks: 0  Help with arithmetic progression question?
Hi, Here's the question: The 4th 5th and 6th terms of an arithmetic series are 2x+1, 3x and x+20. Calculate the sum of the first 10 terms of this series. Here's a picture of my workings: My final answer for Sn for the first 10 terms was 240. Please tell me I'm right, if not, please detail what you did. 
September 20th, 2015, 08:16 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
September 20th, 2015, 11:18 AM  #3  
Senior Member Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177  Quote:
$\displaystyle x=7,\ d=6,\ a_1= 3 $ $\displaystyle d$ is the common difference $\displaystyle a_1$ is the first term The arithmetic progression is : 3, 3, 9, 15, 21, 27, 33, 39, 45, 51, ... $\displaystyle S_{10}=\dfrac{10}{2}(a_1+a_{10})=5(3+51 )=5 \cdot 48=240 .$  
September 20th, 2015, 04:57 PM  #4 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 661 Thanks: 87 
x and d were solved for before me, but what I did was 2x + 1 + x + 20 = 2(3x) because twice the fifth term equals the sum of the fourth and sixth terms since the fifth term is the mean of the fourth and sixth terms. The equation adds up to 3x + 21 = 6x, then 21 = 3x, then x = 7. What is useful is that the sum of the first ten terms is equal to five times the sum of the fifth and sixth (middle two) terms*, so you don't need to calculate all ten terms to find the sum. * For example, the sum of the numbers from 1 to 10 = 55 = 5(5+6) Last edited by EvanJ; September 20th, 2015 at 04:59 PM. 
September 23rd, 2015, 10:34 AM  #5 
Newbie Joined: Sep 2015 From: United Kingdom Posts: 8 Thanks: 0 
Thanks for the help guys 

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