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If the pth, qth & rth term of an AP is x, y and z respectively, show that x(q-r) + y(r-p) + z(p-q) = 0
Ans: pth term ⇒ x = A + (p-1) D
qth term ⇒ y = A + (q-1) D
rth term ⇒ z = A + (r-1) D
T.P x(q-r) + y(r-p) + z(p-q) = 0
={A+(p-1)D}(q-r) + {A + (q-1)D} (r-p)
+ {A+(r-1)D} (p-q)
A {(q-r) + (r-p) + (p-q)} + D {(p-1)(q-r)
+ (r-1) (r-p) + (r-1) (p-q)}
⇒ A.0 + D{p(q-r) + q(r-p) + r (p-q)
- (q-r) - (r-p)-(p-q)}
= A.0 + D.0 = 0.
Hence proved
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in right angle triangle BAC.
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