Question

If a, b and c are pth, qth and rth term of an AP
Prove that :
a(q - r) + b(r - p) + c(p - q) = 0

Full Explanation needed! ✌️​

Answer 1

Let e be first term and d be common difference

a = e + ( p-1) d..... (1)

b = e + ( q-1) d......(2)

c= e + ( r -1)d......(3)

(1) - (2)


a - b= d ( p-q)

p- q = a - b)/d

(2) -(3)

b -c = d( q-r)

q - r= b-c)/d

(1) - (3)

a- c= d( p-r)

p -r = a-c)/d

Now

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

substitute

a( b-c)/d + b( c-a)/d + c( a-b)/d

ab - ac + bc - ba + ca - cb)/d

0

✌✌✌✌✌Dhruv✌✌✌✌

Answer 2

We know that the n th term of an A.P is given by a₁ + ( n - 1 ) d

where a₁ = first term .

n = number of terms .

d = common difference .

p th term will be a₁ + ( p - 1 ) d

q th term will be a₁ + ( q - 1 ) d

r th term will be a₁ + ( r - 1 ) d

We can write the above things as :

a = a₁ + ( p - 1 ) d .......( 1 )

b = a₁ + ( q - 1 ) d .......( 2 )

c = a₁ + ( r - 1 ) d ......( 3 )

Subtract ( 2 ) from ( 1 )  :

a - b = ( p - 1 ) d - ( q - 1 ) d

Take d as common :

⇒ a - b = d ( p - 1 - q + 1 )

⇒ a - b = d ( p - q ) ......( 4 )

Subtract ( 3 ) from ( 2 ) :

b - c = ( q - 1 ) d - ( r - 1 ) d

Take d as common :

⇒ b - c = d ( q - 1 - r + 1 )

⇒ b - c = d ( q - r ) ......( 5 )

Subtract ( 1 ) from ( 3 ) :

c - a = ( r - 1 ) d - ( p - 1 ) d

Take d as common :

⇒ c - a = d ( r - 1 - p + 1 )

⇒ c - a = d ( r - p ) ......( 5 )

From 4 , 5 and 6 we can write :

p - q = ( a - b )/d .....( 7 )

q - r = ( b - c )/d ....( 8 )

r - p = ( c - a )/d .....( 9 )

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

⇒ a ( b - c )/d + b ( c - a )/d + c ( a - b )/d

⇒ ( ab - ac )/d + ( bc - ab )/d + ( ac - bc )/d

⇒ ( ab - ac + bc - ab + ac - bc )/d

⇒ 0/d

⇒ 0

L.H.S = R.H.S

Hence it is proved !