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 June 15th, 2018, 09:40 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, 09:49 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,820 Thanks: 1464 $a_1 + (a_1+d) + (a_1+2d) + (a_1+3d) + (a_1+4d) = 40$ $a_1^2 + (a_1+d)^2 + (a_1+2d)^2 + (a_1+3d)^2 + (a_1+4d)^2 = 410$ solve the system for $a_1$ and $d$ Thanks from romsek and Chanceux
June 15th, 2018, 09:52 AM   #3
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Quote:
 Originally Posted by Chanceux The sum of five numbers in arithmetic progression is 40
$k+(k+m)+(k+2m)+(k+3m)+(k+4m) = 40$

Quote:
 and the sum of their squares is 410.
$k^2+(k+m)^2+(k+2m)^2+(k+3m)^2+(k+4m)^2 = 410$

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 10:27 AM.

 June 15th, 2018, 09:59 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 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. Thanks from Chanceux
 June 15th, 2018, 10:38 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,310 Thanks: 1981 $a^2 + 4ad + 6d^2 - (a + 2d)^2 = 82 - 8^2$, 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. Thanks from Chanceux
 June 15th, 2018, 11:59 AM #6 Newbie   Joined: Jun 2018 From: IIds Posts: 2 Thanks: 1 Great advice from all of you, guys! Big thanks to everyone. Thanks from MarkFL
 June 15th, 2018, 01:06 PM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,988 Thanks: 995 Hey Chanceux, are you Lucky ?
 June 15th, 2018, 03:27 PM #8 Math Team   Joined: Jul 2011 From: Texas Posts: 2,820 Thanks: 1464 $(a_3-2d) + (a_3-d) + a_3 + (a_3+d) + (a_3+2d) = 40$ $5a_3 = 40 \implies a_3 = 8$ $(a_3-2d)^2 + (a_3-d)^2 + a_3^2 + (a_3+d)^2 + (a_3+2d)^2 = 410$ $(8-2d)^2 + (8-d)^2 + 8^2 + (8+d)^2 + (8+2d)^2 = 410$ $(64 - 32d + 4d^2) + (64 - 16d + d^2) + 64 + (64+16d + d^2) +(64+32d+4d^2) = 410$ $320 + 10d^2 = 410$ $d^2 = 9 \implies d = \pm 3$ for $d = 3$ ... $\{ 2,5,8,11,14 \}$ for $d = -3$ ... $\{14,11,8,5,2 \}$

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