Question

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

Answer 1

Pls. Find the answer

Hope it helps you

Answer 2

ANSWER :

• Let, A be the first term
• D be the common difference

• In this question we have to use this ( Sn ) formula :

       Sn = n/2 { 2a + ( n -1 ) d }

a = sum of p terms.

⇒ a = p/2 { 2A + ( p - 1 ) D }

⇒ 2a/p = { 2A + ( p - 1 ) D }                ------------  ( i )

b = sum of q terms.

⇒ b = q/2 { 2A + ( q - 1 } D

⇒ 2b / q = { 2A + ( q - 1 ) D }            ------------- ( ii )

And now,   c = sum of r terms.

⇒ c = r/2 { 2A + ( r - 1 ) D }

⇒ 2c/r = { 2A + ( r - 1 ) D }               ------------- ( iii )

Multiply eq ( i ), ( ii ) and ( iii ) by (q-r) , (r-p) and (p-q) and add we get :

2a/p ( q-r ) + 2b/q ( r-p ) + 2c/r ( p - q )

{ 2A + ( p-1 ) D } ( q - r ) + { 2A + ( q-1 ) D } ( r - p ) + { 2A + ( r - p ) D } ( p - q )

2A {( q - r + r - p + p - q ) + D { ( p - 1 ) ( q - r ) + ( q - 1 ) ( r - p ) + ( r - 1 ) ( p - q )}

= 2A × 0 + D × 0 = 0

HENCE IT IS PROVED...!!!