Question

If pth , qth and rth term of an AP are a,b,c respectively , then show that

(a-b)r +(b-c)p + (c-a)q = 0



Please help me with this question asap .

Answer 1

Heya User,

T(p) = a  ||  T(q) = b  ||   T(r) = c 

--> T(1) + ( p - 1 )d = a
--> T(1) + ( q - 1 )d = b
--> T(1) + ( r - 1 )d = c 

--> ( a - b ) = ( p - q )d
--> ( b - c ) = ( q - r )d
--> ( c - a ) = ( r - p )d

--> ( a - b )r + ( b - c )p + ( c - a )q  = d [ ( p - q )r + ( q - r )p + ( r - p )q ]
                              = d [ pr - qr + pq - pr + qr - pq ] = d ( 0 ) = 0

Hence, ( a - b )r + ( b - c )p + ( c - a )q = 0

Answer 2

LHS=RHS $(a-b)r +(b-c)p + (c-a)q=0$

Step-by-step explanation:

To show : $(a-b)r +(b-c)p + (c-a)q=0$

In an A.P the nth term with first term a' and common difference d is given  by, $a_n=a'+(n-1)d$

The pth term of A.P is a,

$a=a'+(p-1)d$

The qth term of A.P is b,

$b=a'+(q-1)d$

The rth term of A.P is c,

$c=a'+(r-1)d$

Now,

$r(a-b)=r[a'+(p-1)d-(a'+(q-1)d)]$

$r(a-b)=r(p-q)d$ ......(1)

$p(b-c)=p[a'+(q-1)d-(a'+(r-1)d)]$

$p(b-c)=p(q-r)d$ .....(2)

$q(c-a)=q[a'+(r-1)d-(a'+(p-1)d)]$

$q(c-a)=q(r-p)d$ ......(3)

Substitute the value of (1), (2), (3) in LHS of expression

$LHS=(a-b)r +(b-c)p + (c-a)q$

$=r(p-q)d+p(q-r)d+q(r-p)d$

$=d(pq-pr+qr-pq+rp-qr)$

$=d(0)$

$=0$

=RHS

#Learn more

In an AP pth, qth and rth terms are respectively a, b and c. Prove that

p(b - c) + q(c - a) + r(a - b) = 0

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