A particle in uniformly accelerated motion travels a, b and c distances in xth, yth, and zth second of its motion, respectively. Then a(y - z) + b(z - x) + c (x - y) =
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Using S n + u + 
(2n - 1) we get
a = u + 
(2x - 1) …………… (i)
[d = uniform acceleration]
b = u + 
(2y - 1)…………… (ii)
c = u + 
(2z - 1) …………… (iii)
or a = dx + 
…………….. (iv)
From (iv) ay = dxy + 
y
and az = dxy + 
z
Subtracting, a(y - z)
= d(xy - xz) + 
(y - z)
Similarly, b(z - x)
= a'(yz – yx) + 
(x - y)
and c(x - y) = d(zx - yz) + 
(x - y)
Adding above 3 equations
a(y - z) + b(z - x) + c(x - y)
= a'(xy – xz + yz – yx + xz - yz) + 
(y – z + z – x + x - y) = 0
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