Question

if p,q and r term of an ap is x,y,and z respectively show that x(q-r) +y(r-p)+z(p-q) is equal to zero ​

Answer 1

Step-by-step explanation:

→ Given:-

Pth term = x.

Qth term = y.

And, Rth term = z.

→ To prove :-

⇒ x( q - r ) + y( r - p ) + z( p - q ) = 0.

→ Solution:-

Let a be the first term and D be the common difference of the given AP. Then,

$T_p$ = a + ( p - 1 )d.

$T_q$ = a + ( q - 1 )d.

And,

$T_r$ = a + ( r - 1 )d.

▶ Now,

⇒ a + ( p - 1 )d = x..........(1).

⇒ a + ( q - 1 )d = y..........(2).

⇒ a + ( r - 1 )d = z...........(3).

▶ On multiplying equation (1) by ( q - r ), (2) by ( r - p ) and (3) by ( p - q ), and adding, we get

⇒ x( q - r ) + y( r - p ) + z( p - q ) = x•{( q - r ) + ( r - p ) + ( p - q )} + d•{( p - 1 ) ( q - r ) + ( q - 1 ) ( r - p ) + ( r - 1 ) ( p - q )}

⇒ x( q - r ) + y( r - p ) + z( p - q ) = ( x × 0 ) + ( d × 0 ).

⇒ x( q - r ) + y( r - p ) + z( p - q ) = 0.

Hence, it is proved .