Question

Sum of the first p,q and r
terms of an AP are a, b and c
respectively. Prove that
a/p(q-r)+ b/q (r-p) +c/r (P-q)=0
​

Answer 1

Answer:

Step-by-step explanation:

Let the first term is a and common difference is d .

//We know that Sum of n terms of an A.P. = n/2[2a + (n - 1)d]

Sum of first p terms, Sp = a  

p/2 [2a + (p - 1) d] = a

=> a/p = 1/2[2a + (p - 1)d]   ----- (1)

Sum of first q terms, Sq = b

q/2 [2a + (q - 1) d] = b  

=> b/q = 1/2[2a + (q - 1)d]   ------ (2)

Sum of first r terms, Sr = c

r/2[2a + (r - 1) d] = c  

=> c/r = 1/2[2a+ (r - 1)d] ------ (3)

Now L.H.S

a/p * (q - r) + b/q * (r - p) + c/r * (p - q)

=> 1/2[2a + (p - 1)d] * (q - r)  + 1/2[2a + (q - 1)d]  * (r - p) + 1/2[2a+ (r - 1)d]  * (p - q)

=> 1/2 [ 2a(q - r) + 2a (r - p) + 2a(p - q) + d [(p - 1)( q - r) + d ( q - 1)(r - p) + d ( r - 1)(p - q)]

=> 1/2 [ 2a(q - r + r - p + p - q) + d (pq - pr - q + r + qr - qp - r + p + rp - rq - p +q)]

=> 1/2 [ 2a (0) + d (0)]

=> 0

= R.H.S

Hence proved.