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

##### 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

### Related Questions

#### Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

#### Find the two numbers whose A.M. is 25 and GM is 20.

Find the two numbers whose A.M. is 25 and GM is 20.

#### If a is the G.M. of 2 and 1/4 find a.

If a is the G.M. of 2 and $$\dfrac{1}{4}$$ find a.

#### Find the geometric means of the following pairs of numbers

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a3b and ab3

(iii) –8 and –2

Insert 5 geometric means between $$\dfrac{32}{9}$$ and $$\dfrac{81}{2}$$.