if a,b,c be respectively the Xth, Yth and Zth terms of an AP prove that a(y-z) + b(z-x) + c( x-y)=0
Answers
Step-by-step explanation :
Given,
a,b,c be respectively the xth, yth and zth terms of an AP
To prove,
a(y - z) + b(z - x) + c(x - y) = 0
Solution,
nth term of an AP is given by,
tₙ = t + (n - 1)d
where
t is the first term
d is the common difference
- xth term of AP = a
a = t + (x - 1)d --[1]
- yth term of AP = b
b = t + (y - 1)d --[2]
- zth term of AP = c
c = t + (z - 1)d --[3]
equation [1] - equation [2]
a - b = t + (x - 1)d - [t + (y - 1)d]
a - b = t + xd - d - [t + yd - d]
a - b = t + xd - d - t - yd + d
a - b = xd - yd
equation [2] - equation [3]
b - c = t + (y - 1)d - [t + (z - 1)d]
b - c = t + yd - d - [t + zd - d]
b - c = t + yd - d - t - zd + d
b - c = yd - zd
equation [3] - equation [1]
c - a = t + (z - 1)d - [t + (x - 1)d]
c - a = t + zd - d - [t + zx - d]
c - a = t + zd - d - t - zx + d
c - a = zd - xd
we have to prove a(y - z) + b(z - x) + c(x - y) = 0
LHS = a(y - z) + b(z - x) + c(x - y)
= ay - az + bz - bx + cx - cy
= ay - cy - az + bz - bx + cx
= y(a - c) + z(b - a) + x(c - b)
= -y(c - a) - z (a - b) - x (b - c)
= -y (zd - xd) - z(xd - yd) - x(yd - zd)
= -yzd + xyd - zxd + yzd - xyd + zxd
= -yzd + yzd + xyd - xyd + zxd - zxd
= 0
= RHS
∴ LHS = RHS
a(y - z) + b(z - x) + c(x - y) = 0
Hence proved!