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Question: If the sum of first *p* terms, first *q* terms and first *r* terms of an A.P. be *x*, *y* and *z* re...

If the sum of first p terms, first q terms and first r terms of an A.P. be x, y and z respectively, then xp(q−r)+yq(r−p)+zr(p−q)\frac{x}{p}(q - r) + \frac{y}{q}(r - p) + \frac{z}{r}(p - q) is

A

0

B

2

C

pqr

D

8xyzpqr\frac{8xyz}{pqr}

Answer

0

Explanation

Solution

We have a, the first term and d, the common difference, x={2a+(p−1)d}p2x = \{ 2a + (p - 1)d\}\frac{p}{2} ⇒ xp=a+(p−1)d2\frac{x}{p} = a + (p - 1)\frac{d}{2}

Similarly, yq=a+(q−1)d2\frac{y}{q} = a + (q - 1)\frac{d}{2} and zr=a+(r−1)d2\frac{z}{r} = a + (r - 1)\frac{d}{2}

∴ xp(q−r)+yq(r−p)+zr(p−q)={a+(p−1)d2}(q−r)+{a+(q−1)d2}(r−p)+{a+(r−1)d2}(p−q)\frac{x}{p}(q - r) + \frac{y}{q}(r - p) + \frac{z}{r}(p - q) = \left\{ a + (p - 1)\frac{d}{2} \right\}(q - r) + \left\{ a + (q - 1)\frac{d}{2} \right\}(r - p) + \left\{ a + (r - 1)\frac{d}{2} \right\}(p - q) =a{(q−r)+(r−p)+(p−q)}+d2{(p−1)(q−r)+(q−1)(r−p)+(r−1)(p−q)}= a\{(q - r) + (r - p) + (p - q)\} + \frac{d}{2}\{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)\}

=a..0+d2[{pq−pr+rq−pq+pr−qr−{(q−r)+(r−p)+(p−q)}=0+d2{0−0}=0= a..0 + \frac{d}{2}\lbrack\{ pq - pr + rq - pq + pr - qr - \{(q - r) + (r - p) + (p - q)\} = 0 + \frac{d}{2}\{ 0 - 0\} = 0