Question

If lth,mth,nth terms of an ap are x,y,z respectively,then show that x(m-n)+y(n-l)+z(l-m)=0

Answer 1

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▪️→ 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,

▪️Tp= a + ( p - 1 )d.

▪️Tp = a + ( q - 1 )d.

▪️And,

▪️Tr= 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

Proved

Answer 2

▪️→ 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,

▪️Tp= a + ( p - 1 )d.

▪️Tp = a + ( q - 1 )d.

▪️And,

▪️Tr= 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

Proved