Question

Do fasttt....................​

Answer 1

Ah yes, sequences and series my forte.

→ 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 = a + ( p - 1 )d.

T = a + ( q - 1 )d.

And,

T = 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.

✔✔ QED, Boom!

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THANKS#ImSoLonely

11th std is hard.