Question

the sum of the first p, q, r terms of an AP are a, b, c respectively. prove that
a/p(q-r) + b/q(r-p) + c/r(p-q) =0​

Answer 1

$\afrac{a}{p}$· (q - r) + $\afrac{b}{q}$· (r - p) + $\afrac{c}{r}$· (p - q) = 0, is is proved.

Step-by-step explanation:

Let first term = x and common difference = d

∴ $S_{p}$ = $\dfrac{p}{2}$·{2x + (p - 1)d}

⇒ a = $\dfrac{p}{2}$·{2x + (p - 1)d}

⇒ $\afrac{a}{p}$ = x + (p -1)·$\dfrac{d}{2}$    ..... (1)

Similarly,

$\afrac{b}{q}$ = x + (q -1)·$\dfrac{d}{2}$       ..... (2)

$\afrac{c}{r}$ = x + (r -1)·$\dfrac{d}{2}$          ..... (3)

Multipying (1) × (q - r), (2) × (r - p) and (3) × (p - q) and adding them, we get

$\afrac{a}{p}$· (q - r) + $\afrac{b}{q}$· (r - p) + $\afrac{c}{r}$· (p - q) = x·(q - r + r - p + p - q) + tex]\dfrac{d}{2}[/tex]·{(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q) }  

= x·(0) + tex]\dfrac{d}{2}[/tex]·{pq - pr - q + r + qr - qp - r + p + rp - rq - p + q }  

= x·(q - r + r - p + p - q) + tex]\dfrac{d}{2}[/tex]· (0)

= 0

Hence, it is proved.