Subjective Type

Sum of the first $$p, q$$ and $$r$$ terms of an A.P. are $$a,b$$ and $$c$$ respectively. Prove that $$\displaystyle \frac { a }{ p } (q-r)+\frac { b }{ q } (r-p)+\frac { c }{ r } (p-q)=0$$

Solution

Sum of $$n$$ terms of an A.P. is given by
$$S_n =\cfrac{n}{2} [2a+(n-1)d]$$

Let $$\displaystyle {a}_{1}$$ and $$d$$ be the first term and common difference of A.P. respectively.

Given, $$S_p =a$$

$$\Rightarrow \displaystyle \cfrac { p }{ 2 } \left[ { 2a }_{ 1 }+(p-1)d \right] =a$$

$$ \Rightarrow { 2a }_{ 1 }+(p-1)d=\cfrac { 2a }{ p } $$ ...(1)

Also, given $${ S }_{ q }=b$$

$$\Rightarrow \cfrac { q }{ 2 } \left[ { 2a }_{ 1 }+(q-1)d \right] =b$$

$$\Rightarrow { 2a }_{ 1 }+(q-1)d=\cfrac { 2b }{ q }$$ ...(2)

Also given $$S_r=c$$

$$\Rightarrow \displaystyle \frac { r }{ 2 } \left[ { 2a }_{ 1 }+(r-1)d \right] =c$$

$$ \Rightarrow \displaystyle { 2a }_{ 1 }+(r-1)d=\frac { 2c }{ r } $$ ...(3)

Subtracting (2) from (1), we get

$$(p-1)d-(q-1)d\quad =\displaystyle \frac { 2a }{ p } -\frac { 2b }{ q } $$

$$\Rightarrow d(p-1-q+1)=\cfrac { 2aq-2bq }{ pq }$$

$$\Rightarrow d(p-q)=\cfrac { 2aq-2bp }{ pq } $$

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

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

$$ (q-1)d-(r-1)d\quad =\displaystyle \frac { 2b }{ q } -\frac { 2c }{ r } $$

$$ \Rightarrow d(q-1-r+1)=\cfrac { 2b }{ q } -\cfrac { 2c }{ r } $$

$$ \Rightarrow d(q-r)=\cfrac { 2br-2qc }{ qr } $$

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

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

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

$$\Rightarrow qr=(q-r)(aq-bq)=pq(p-q)(br-qc)$$

$$ \Rightarrow r=(aq-bp)(q-r)=p(br-qc)(p-q)$$

$$ \Rightarrow (aqr-bpr)(q-r)=(bpr-pqc)(p-q)$$ Dividing both sides by $$pqr$$,

we get

$$\displaystyle \left( \frac { a }{ p } -\frac { b }{ q } \right) (q-r)=\left( \frac { b }{ q } -\frac { c }{ r } \right) (p-q)$$

$$ \Rightarrow \displaystyle \frac { a }{ p } (q-r)-\frac { b }{ q } (q-r+p-q)+\frac { c }{ r } (p-q)=0$$

$$\Rightarrow \displaystyle \frac { a }{ p } (q-r)+\frac { b }{ q } (r-p)+\frac { c }{ r } (p-q)=0$$ $$[hence \, proved]$$

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