Question

If pth, qth, and rth terms of an AP are a,b,c respectively,then show that.
a(q-r)+b(r-p)+c(p-q)=0

Answer 1

Answer:-

Given:

pth , qth , rth terms of an AP are a , b , c.

We know that,

nth term of an AP (a(n)) = a + (n - 1)d

Hence,

→ a + (p - 1)d = a -- equation (1)

→ a + (q - 1)d = b -- equation (2)

→ a + (r - 1)d = c -- equation (3)

Using common difference,

→ eq(3) - eq(2) = eq(2) - eq(1)

→ a + (r - 1)d - [a + (q - 1)d] = a + (q - 1)d - [a + (p - 1)d]

→ a + rd - d -( a + qd - d) = a + qd - d - (a +

pd - d)

→ a + rd - d - a - qd + d = a + qd - d - a - pd + d

→ rd - qd = qd - pd

→ (r - q) * d = (q - p) * d

→ r - q = q - p

Hence, p , q , r are also in AP since there is common difference between three terms.

→ q - r = p - q. ( 2q = p + r) => q = (p + r)/2

→ c - b = b - a. (2b = a + c) => b = (a + c)/2

We have to prove:

a(q - r) + b(r - p) + c (p - q) = 0

Substitute the value of "b" here.

$\sf \implies \large{ aq - ar + ( \frac{(a + c)(r - p)}{2} ) + cp - cq = 0}$

$\sf \implies \large{ aq - ar + ( \frac{ar - ap + cr - cp}{2} ) + cp - cq = 0}$

Taking LCM in LHS we get,

$\sf \implies \large{ \frac{2aq - 2ar + ar + cr - ap - cp + 2cp - 2cq}{2} = 0}$

$\sf \implies \large{ \frac{2aq - 2cq - ar - ap + cr + cp}{2} = 0}$

$\sf \implies \large{\frac{2aq - 2cq - a(r + p) + c(p + r)}{2} = 0}$

Putting the value of (p + r) as 2q we get,

$\sf \implies\large{ \frac{2aq - 2cq - a(2q) + c(2q)}{2} = 0}$

$\sf \implies \large{ \frac{2aq - 2cq - 2aq + 2cq}{2} = 0}$

$\sf \implies \: \frac{0}{2} = 0$

$\implies \sf\large{0 = 0}$

→ LHS = RHS

Hence, Proved.