The sum of the four terms in arithmetic progression is 18 and the sum of their squares is 326. find the number?


Sagot :

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

So we know that:
Let us make[tex]a_1=m, a_2=n, a_3=o, a_4=p[/tex]
[tex]m+n+o+p=18[/tex]
[tex]m^2+n^2+o^2+p^2=326[/tex]

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

[tex](n-d)+n+(n+d)+(n+2d)=4n+2d=18 \\ n= \frac{9-d}{2} [/tex]

We plug in the values to the second equation in the given
[tex] (\frac{9-3d}{2}) ^2+ (\frac{9-d}{2}) ^2+ (\frac{9+d}{2}) ^2+ (\frac{9+3d}{2}) ^2=326[/tex]
[tex]\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[/tex]

The squareroot of 49 is either positive or negative 7.

So remember that
[tex]n= \frac{9-d}{2} [/tex]

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

If d=-7
[tex]n= \frac{9-(-7)}{2} = \frac{16}{2} =8[/tex]
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