SOLUTION: if p,q,r,s are any four consecutive terms of an A.P. Show that p�-3q�+3r�-s� = 0

Question 696761: if p,q,r,s are any four consecutive terms of an A.P. Show that
p�-3q�+3r�-s� = 0
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!

If d is the common difference in the A.P., then

q = p+d,  r = p+2d,  s = p+3d

p�-3q�+3r�-s� = 

p� - 3(q�-r�) - s� =

(p�-s�) - 3(q�-r�) =

Factor both parentheses as the difference of squares:

[p-s][p+s] - 3[q-r][q+r] =

Substitute q = p+d,  r = p+2d,  s = p+3d

[p-(p+3d)][p+(p+3d)] - 3[(p+d)-(p+2d)][(p+d)+(p+2d)] =

[p-p-3d][p+p+3d] - 3[p+d-p-2d][p+d+p+2d] =

[-3d][2p+3d] - 3[-d][2p+3d] =

-3d[2p+3d] + 3d[2p+3d] =

-6dp -9d� + 6dp + 9d� =
   
          0

Edwin

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