Question

Hello friends ,

Plz solve this question,

If the Pth, Qth and Rth terms of an AP be a,b,c respectively then show that

a(p-r) +b(r-p)+c(p-q)= 0.

Answer 1

Hey there !!


→ Given:-

Pth term = a.
Qth term = b.
And, Rth term = c.

→ To prove :-

=> a( q - r ) + b( r - p ) + c( p - q ) = 0.

→ Solution:-


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

$T \tiny{p}$ = x + ( p - 1 )d.

$T \tiny{q}$ = x + ( q - 1 )d.

And,

$T \tiny{r}$ = x + ( r - 1 )d.


▶ Now,

=> x + ( p - 1 )d = a..........(1).

=> x + ( q - 1 )d = b..........(2).

=> x + ( r - 1 )d = c...........(3).


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

=> a( q - r ) + b( r - p ) + c( p - q ) = x•{( q - r ) + ( r - p ) + ( p - q )} + d•{( p - 1 ) ( q - r ) + ( q - 1 ) ( r - p ) + ( r - 1 ) ( p - q )}


=> a( q - r ) + b( r - p ) + c( p - q ) = ( x × 0 ) + ( d × 0 ).

$\boxed{ \bf => a( q - r ) + b( r - p ) + c( p - q ) = 0. }$


✔✔ Hence, it is proved ✅✅.

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Answer 2

Heya!!

I think ur question is lyk tat :

If the pth, qth and rth terms of an AP be a,b,c respectively then show that

a(q-r) +b(r-p)+c(p-q)= 0 :

Since pth term of an AP = a

=) X + (p-1)D = a .. Eq1.

Here X : first no,
D : common diff

Similarly :

=) X + (q-1)D = b .. Eq2.

&

=) X + (r-1)D = c .. Eq3.

To find :


a(q-r) +b(r-p)+c(p-q)= 0.

Multiply eq 1 by q-r, eq 2 by r-p & eq3 by p-q:

=) (X+ pD - D)(q-r) = a(q-r) .. Eq3.

=) (X+ qD - D) (r-p) = b(r-p) .. Eq4.

=) (X + rD - D) (p-q) = c(p-q) .. Eq5.

Add all eqs ;

=) a(q-r) + b(r-p) +c(p-q) = Xq - Xr + pDq - pDr - Dq + Dr + Xr - Xp + qDr - qDp - Dr + Dp + Xp - Xq + rDp - rDq - Dp + Dq

=) a(q-r) + b(r-p) +c(p-q) = 0

Hope it helps uh!