Question

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

SHOW THAT :

$\frac{a}{p} (q - r) + \frac{b}{q} ( r - p) + \frac{c}{r}(p - q) = 0$
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BEST ANSWER WILL BE MARKED AS BRAINLIEST! ✔️

Answer 1

Note: Sum of n terms = (n/2)[2a + (n - 1) * d].

Let the first term be A and the common difference be d.

(i)

Given sum of p terms = a.

⇒ (p/2)[2A + (p - 1) * d] = a

⇒ (a/p) = (1/2)[2A + (p - 1) * d]

(ii)

Given sum of q terms is b.

⇒ (q/2) = 2A + (q - 1) * d = b

⇒ (q/p) = (1/2)[2A + (q - 1) * d]

(iii)

Given sum of r terms of an AP is c.

⇒ (r/2)[2A + (r - 1) * d] = c

⇒ (r/c) = (1/2)[2A + (r - 1) * d]

(iv)

On subtracting (ii) from (i), we get

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

$=>\frac{a}{p}-\frac{b}{q}=\frac{1}{2}[(p - q) *d]$

$=> p-q=\frac{2}{d}[\frac{a}{p}-\frac{b}{q}]$

(v)

On subtracting (iii) from (ii), we get

$=>\frac{b}{q}-\frac{c}{r}=\frac{1}{2}[(q-1)d-(r-1)d]$

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

$=>q-r=\frac{2}{d}[ \frac{b}{q}-\frac{c}{r}]$

(vi)

on subtracting (iii) from (i), we get

$=>\frac{c}{r}-\frac{a}{p}=\frac{1}{2}(r-p)d$

$=>r-p =\frac{2}{d}[ \frac{c}{r}- \frac{a}{p}]$

Now,

Given Equation is (a/p)(q - r) + (b/q)(r - p) + (c/r)(p - q) = 0

$=>\frac{a}{q}[\frac{2}{d}(\frac{b}{q}- \frac{c}{r})]+\frac{b}{q}[\frac{2}{d}(\frac{c}{r}-\frac{a}{p})]+\frac{c}{r}[\frac{2}{d}( \frac{a}{p}-\frac{b}{q})]$

$=>\frac{2}{d}[\frac{ab}{pq}- \frac{ac}{pr}+\frac{bc}{qr}-\frac{ab}{pq}+\frac{ac}{pr}-\frac{bc}{qr}]$

$=>0.$

Hope it helps!