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June 15th, 2018, 08:40 AM  #1 
Newbie Joined: Jun 2018 From: IIds Posts: 2 Thanks: 1  Five numbers in arithmetic series
The sum of five numbers in arithmetic progression is 40 and the sum of their squares is 410. Find the five numbers. Thanks beforehand for showing the way to solution 
June 15th, 2018, 08:49 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,756 Thanks: 1407 
$a_1 + (a_1+d) + (a_1+2d) + (a_1+3d) + (a_1+4d) = 40$ $a_1^2 + (a_1+d)^2 + (a_1+2d)^2 + (a_1+3d)^2 + (a_1+4d)^2 = 410$ solve the system for $a_1$ and $d$ 
June 15th, 2018, 08:52 AM  #3  
Senior Member Joined: Sep 2015 From: USA Posts: 2,011 Thanks: 1044  $k+(k+m)+(k+2m)+(k+3m)+(k+4m) = 40$ Quote:
Expand all this out into a linear equation and a quadratic equation which will have two solutions. Last edited by skipjack; June 15th, 2018 at 09:27 AM.  
June 15th, 2018, 08:59 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
I would begin by writing: $\displaystyle (a)+(a+d)+(a+2d)+(a+3d)+(a+4d)=40$ $\displaystyle 5a+10d=40$ $\displaystyle a+2d=8$ Then: $\displaystyle (a)^2+(a+d)^2+(a+2d)^2+(a+3d)^2+(a+4d)^2=410$ This reduces to $\displaystyle a^2+4ad+6d^2=82$ Now you have two equations and two unknowns...can you proceed? edit: There were no replies when I started...sorry to be redundant. 
June 15th, 2018, 09:38 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,191 Thanks: 1649 
$a^2 + 4ad + 6d^2  (a + 2d)^2 = 82  8^2$, so $2d^2 = 18$. It follows that the five numbers are 2, 5, 8, 11 and 14. The progression may consist of these numbers in ascending order or in descending order. 
June 15th, 2018, 10:59 AM  #6 
Newbie Joined: Jun 2018 From: IIds Posts: 2 Thanks: 1 
Great advice from all of you, guys! Big thanks to everyone.

June 15th, 2018, 12:06 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,771 Thanks: 862 
Hey Chanceux, are you Lucky ? 
June 15th, 2018, 02:27 PM  #8 
Math Team Joined: Jul 2011 From: Texas Posts: 2,756 Thanks: 1407 
$(a_32d) + (a_3d) + a_3 + (a_3+d) + (a_3+2d) = 40$ $5a_3 = 40 \implies a_3 = 8$ $(a_32d)^2 + (a_3d)^2 + a_3^2 + (a_3+d)^2 + (a_3+2d)^2 = 410$ $(82d)^2 + (8d)^2 + 8^2 + (8+d)^2 + (8+2d)^2 = 410$ $(64  32d + 4d^2) + (64  16d + d^2) + 64 + (64+16d + d^2) +(64+32d+4d^2) = 410$ $320 + 10d^2 = 410$ $d^2 = 9 \implies d = \pm 3$ for $d = 3$ ... $\{ 2,5,8,11,14 \}$ for $d = 3$ ... $\{14,11,8,5,2 \}$ 

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