Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.

Prove that  \( \dfrac{a}{p}(q-r)+\dfrac{b}{q}(r-p)+\dfrac{c}{r}(p-q)=0\)

Asked by Pragya Singh | 1 year ago |  115

1 Answer

Solution :-

Let a1 be the first term and d be the common difference of the A.P.
According to the conditions given in the question,

\( s_p=\dfrac{p}{2}[(2a_1+(p-1)d]\) = a

2a1+(p-1)d = \(\dfrac{2a}{p}\)

 \( s_q=\dfrac{q}{2}[(2a_1+(q-1)d]\) = b

 

2a1+(p-1)d = \(\dfrac{2b}{q}\)

\( s_r=\dfrac{q}{2}[(2a_1+(q-1)d]=c\)

2a1+(r-1)d =\( \dfrac{2c}{r}\)

Subtracting (2) from (1), we get 

(p−1)d−(q−1)d = \( \dfrac{2a}{p}- \dfrac{2b}{q}\)

d(p−1−q+1) = \(\)\( \dfrac{2aq-2bq}{pq}\)

d(p−q) = \( \dfrac{2aq-2bq}{pq}\)

d = \( \dfrac{2(aq-bp)}{pq(p-q)}\)..............(4)

Subtracting eq (3) from eq(2), we get 

(q−1)d−(r−1)d=\( \dfrac{2b}{q}- \dfrac{2c}{r}\)

d(q−1−r+1) = \( \dfrac{2b}{q}- \dfrac{2c}{r}\)

d(q−r)= \( \dfrac{2br-2qc}{qr}\)

\( d= \dfrac{2(br-qc)}{qr(q-r)}\)...........(5)

Equating both the values of d in Eqn (4) and Eqn (5), we get

\( \dfrac{(aq-bp)}{pq(p-q)}= \dfrac{(br-qc)}{qr-(q-r)}\)

qr=(q−r)(aq−bq)=pq(p−q)(br−qc)

r=(aq−bp)(q−r)=p(br−qc)(p−q)

(aqr−bpr)(q−r)=(bpr−pqc)(p−q)  Dividing both sides by  pqr, 

we get 

Answered by Abhisek | 1 year ago

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