Question:
Please answer the question : sum of the first p,q & r terms are ab&c respectively then prove that a/p(q-r)+b/q(r-p)+c/r(p-q)=0
Answer:
Let a is the first term and d is the common difference of the AP
Noe, sum of n terms Sn = (n/2)*{2a + (n - 1)d}
Given, Sp = a
=> (p/2)*{2a + (p - 1)d} = a
=> a/p = {2a + (p - 1)d}/2 .............1
Again, Sq = b
=> (q/2)*{2a + (q - 1)d} = b
=> b/q = {2a + (q - 1)d}/2 ............2
and Sr = c
=> (r/2)*{2a + (r - 1)d} = c
=> c/r = {2a + (r - 1)d}/2 .............3
Now, (a/p)*(q - r) + (b/q)*(r - p) + (c/r)*(p - q)
= [{2a + (p - 1)d}/2]*(q - r) + [{2a + (q - 1)d}/2]*(r - p) + [{2a + (r - 1)d}/2]*(p - q)
= {(q - r + r - p + p - q)*2a}/2 + {(p - 1)*(q - r) + (q - 1)*(r - p) + (r - 1)*(p - q)}*d/2
= {(q - r + r - p + p - q)*2a}/2 + (pq - pr - q + r + qr - qp - r + p + rp - rq - p + q )*d/2
= 0
=> (a/p)*(q - r) + (b/q)*(r - p) + (c/r)*(p - q) = 0
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