Question 696761: if p,q,r,s are any four consecutive terms of an A.P. Show that
p²-3q²+3r²-s² = 0
Answer by Edwin McCravy(20064)   (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
p²-3q²+3r²-s² =
p² - 3(q²-r²) - s² =
(p²-s²) - 3(q²-r²) =
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 -9d² + 6dp + 9d² =
0
Edwin
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