Question

if the pth, qth and rth terms of an ap are x, y and z respectively, prove that:x (q-r)+ y(r-p)+z (p-q)=0

Answer 1

question.if the pth, qth and rth terms of an ap are x, y and z respectively, prove that:x (q-r)+ y(r-p)+z (p-q)=0

Answer.
let A be the first term
d be the difference
then pth term is = A+(p-1)d
x=A+(p-1)d. ..... eq(1)
similarly
Y= A+(q-1)d,.........eq(2)
z=A+(r-1)d................eq(3)

now on subtract
x-y=(p-q)d.......eq( 4)
y-z=(q-r)d..........eq(5)
z-x=(r-p)d............eq(6)

so now multiply eq 1,2,3 respectively by
z,x,y
then z(x-y)=z(p-q)d........eq(7)
x(y-z)=x(q-r)d...............eq(8)
y(z-x)=y(r-p)d......,.........eq(9)
from eq 7, 8, 9 on addition we get

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

Answer 2

Hey there !!

→ 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$\tiny p$ = a + ( p - 1 )d.

T$\tiny q$ = a + ( q - 1 )d.

And,

T$\tiny r$ = 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 ✅✅.

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