Question

The pth, qth and rth terms of an A.P. are a, b and c respectively.

Show that a(q – r) + b(r-p) + c(p – q) = 0​

Answer 1

Solution of your problem. Hope it helps you!

Answer 2

SOLUTION

GIVEN

The pth, qth and rth terms of an A.P. are a, b and c respectively.

TO PROVE

a(q – r) + b(r-p) + c(p – q) = 0

EVALUATION

Let x be the first term and d be the Common Difference of the Arithmetic progression

Therefore

$\sf{}p \: th \: term = x + (p - 1)d = a \: \: .....(1)$

$\sf{}q \: th \: term = x + (q - 1)d = b\: \: .....(2)$

$\sf{}r \: th \: term = x + (r - 1)d = c \: \: .....(3)$

So

$\sf{}a - b$

$\sf{} = x + pd - d - x - qd + d$

$\sf{} = pd - qd$

$\therefore \sf{}a - b = pd - qd \: \: .....(4)$

Similarly

$\sf{}b - c = qd - rd \: \: .....(5)$

$\sf{}c - a = rd - pd \: \: .....(6)$

Now

$\sf{}a(q - r) + b(r - p) + c(p - q)$

$= \sf{}aq - ar + br - bp + cp - cq$

$\sf{} = p(c - b) + q(a - c) + r(b - a)$

$\sf{} = prd - pqd + pqd - qrd + qrd - prd$

$\sf{} = 0$

Hence proved

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