Question 11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that a(q-r) / p + b(r-p) / q + c(p-q) / r = 0
Class X1 - Maths -Sequences and Series Page 185
Answers
Answered by
234
Let the first term is A and common difference is d .
Sp = a
P/2{2A + (P-1)d} = a-----(1)
Sq = b
q/2{2A + (q-1)d} = b------(2)
Sr = c
r/2{2A + (r -1)d} = c------(3)
Now,
LHS = a(q-r)/p + b(r-p)/q + c(p-q)/r
= (q-r)/p ×P/2{2A + (p-1)d} + (r-p)/q ×q/2{2A + (q -1)d} + (p-q)/r × r/2{2A + (r-1)d}
= 1/2 [(q-r)(2A +pd -d) + (r -p)(2A + qd - d)+(p-q)(2A + rd-d)]
= 1/2[2A{(q-r)+(r-p)+(p-q)} +d{p(q-r)+q(r-p)+r(p-q)-d{(q-r)+(r-p)+(p-q)}]
= 1/2[2A{0} +d{pq-pr+qr-qp+rp-rq}-d{0}]
=1/2[0 + 0 + 0]
= 0 = RHS
Sp = a
P/2{2A + (P-1)d} = a-----(1)
Sq = b
q/2{2A + (q-1)d} = b------(2)
Sr = c
r/2{2A + (r -1)d} = c------(3)
Now,
LHS = a(q-r)/p + b(r-p)/q + c(p-q)/r
= (q-r)/p ×P/2{2A + (p-1)d} + (r-p)/q ×q/2{2A + (q -1)d} + (p-q)/r × r/2{2A + (r-1)d}
= 1/2 [(q-r)(2A +pd -d) + (r -p)(2A + qd - d)+(p-q)(2A + rd-d)]
= 1/2[2A{(q-r)+(r-p)+(p-q)} +d{p(q-r)+q(r-p)+r(p-q)-d{(q-r)+(r-p)+(p-q)}]
= 1/2[2A{0} +d{pq-pr+qr-qp+rp-rq}-d{0}]
=1/2[0 + 0 + 0]
= 0 = RHS
Answered by
164
Let the first term of AP be A and common difference be d,
We know that sum if n terms of AP,
Using this formula,
Sum of p term of AP,
Sum of q terms
Similarly,
for Sum of r term of AP,
Adding eq. i) , ii) and iii),
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