Question

If tp=a, tq=b and tr=C are terms of an AP
Then prove that a(q-r) +b(r-p) +C(p-q) =0

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{For \: An \: AP \: \: \: \: t_p = a, t_q=b, t_r = c }$

TO PROVE

$\sf{ a(q - r) + b(r - p) + c(p - q) = 0\: }$

CALCULATION

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

Then

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

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

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

Now

$\sf{b - a }$

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

$\sf{= qd - pd\: }$

So

$\sf{b - a = qd - pd\: }$

Similarly

$\sf{c - b = rd - qd\: }$

$\sf{a- c= pd - rd\: }$

Now

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

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

$= \sf{ p(rd - qd) + q(pd - rd) + r(qd - pd) \: }$

$= \sf{ prd - pqd + pqd - qrd + qrd - p rd\: }$

$\sf{ = 0}$

$\therefore \: \: \sf{ a(q - r) + b(r - p) + c(p - q) = 0\: }$

Hence proved

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