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2015-05-04T20:58:27+08:00

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An arithmetic progression is a sequence in which they have a common difference. A sequence is shown as:
a_1,a_2,a_3,...,a_n
a_2-a_1=a_3-a_2=...=a_n-a_n_-_1=d
in which d is the common difference

So we know that:
Let us makea_1=m, a_2=n, a_3=o, a_4=p
m+n+o+p=18
m^2+n^2+o^2+p^2=326

Remember that this is an arithmetic progression so the terms are respectively 
m, m+d, m+2d, m+3d or
n-d, n, n+d, n+2d
Since d is their common difference

(n-d)+n+(n+d)+(n+2d)=4n+2d=18 \\ n= \frac{9-d}{2}

We plug in the values to the second equation in the given
 (\frac{9-3d}{2}) ^2+  (\frac{9-d}{2}) ^2+ (\frac{9+d}{2}) ^2+ (\frac{9+3d}{2}) ^2=326
\frac{2(9^2+(3d)^2)+2(9^2+d^2)}{4} =326 \\ \frac{2*9^2+9d^2+d^2}{2} =326 \\ 81+5d^2=326 \\ 5d^2=245 \\ d^2=49 \\ d=+7 , -7

The squareroot of 49 is either positive or negative 7.

So remember that
n= \frac{9-d}{2}

If d=7
n= \frac{9-7}{2} =1
When n=1
m=1-7=-6
n=1
o=1+7=8
p=8+7=15

If d=-7
n= \frac{9-(-7)}{2} = \frac{16}{2} =8
When n=8
m=8-(-7)=15
n=8
o=8-7=1
p=1-7=-6

Therefore the numbers are -6, 1, 8 and 15



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