Question

If p q r term of an hp be a,b,c prove that (q-r)bc+(r-p)ac+(p-q)ab=0

Answer 1

Answer:

Step-by-step explanation:

Given :

p , q & r th terms of an H.P are a , b & c.

Then, prove that :

$(q-r)bc+(r-p)ac+(p-q)ab=0$

Solution :

We know that,

$a_n = \frac{1}{a + (n-1)d}$

⇒ $a_p = \frac{1}{A + (p-1)d} = a$

⇒ $A + (p-1)d = \frac{1}{a}$ ...(i)

⇒ $a_q = \frac{1}{A + (q-1)d} = b$

⇒ $A + (q-1)d = \frac{1}{b}$ ...(ii)

⇒ $a_r = \frac{1}{A + (r-1)d} = c$

⇒ $A + (p-1)d = \frac{1}{c}$ ...(iii)

So,

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

⇒ $\frac{c-b}{bc} = A + qd - d - A - rd + d$

⇒ $\frac{c-b}{bc} = (q - r)d$

⇒ $\frac{c-b}{d} = (q -r)bc$

⇒ $\frac{-(b-c)}{d} = (q -r)bc$ ...(iv)

$\frac{1}{a} - \frac{1}{c} = A + (p-1)d - [A + (r - 1)d]$

⇒ $\frac{c-a}{bc} = A + pd - d - A - rd + d$

⇒ $\frac{c - a}{ac} = (p-r)d$

⇒ $\frac{a-c}{d} = (r-p)ac$ ...(v)

$\frac{1}{a} - \frac{1}{b} = A+(p-1)d - [A+(q-1)d]$

⇒ $\frac{b-a}{ab} = A+ pd - d - qd + d$

⇒ $\frac{b-a}{ab} = (p-q)d$

⇒ $\frac{b-a}{d} = (p-q)ab$ ...(vi)

By adding (iv) , (v), (vi),

We get,

⇒  $\frac{-(b-c)}{d} + \frac{a-c}{d} + \frac{b-a}{d} = (q -r)bc + (r-p)ac + (p-q)ab$

⇒ $(q -r)bc + (r-p)ac + (p-q)ab = \frac{ -b + c + a - c+ b - a}{d}$

⇒ $(q -r)bc + (r-p)ac + (p-q)ab = 0$

Hence, proved.