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Question
The sum of the first p,q,r terms of an A.P. are a,b,c respectively.Show that ap(q−r)+bq(r−p)+cr(p−q)=0.
Show that ap(q−r)+bq(r−p)+cr(p−q)=0.
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Solution
let A be the first term and D be the common difference of the given A.P. Then, a= Sum of p terms
⟹a=p2{2A+(p−1)D}
⟹2ap={2A+(p−1)D} ...(i)
b= Sum of q terms
⟹b=q2{2A+(q−1)D}
⟹2bq={2A+(q−1)D} ...(ii)
and, c= Sum of r terms
⟹c=r2{2A+(r−1)D}
⟹2cr={2A+(r−1)D} ...(iii)
Multiplying equation (i), (ii) and (iii) by (q−r),(r−p) and (p−q) respectively and adding, we get
2ap(q−r)+2bq(r−p)+2cr(p−q)
={2A+(p−1)D}(q−r)+{2A+(q−1)D}(r−p)+{2A+(r−1)D}(p−q)
=2A(q−r+r−p+p−q)+D{(p−1)(q−1)(r−p)+(r−1)(p−q)
=2A×0+D×0=0
Then, a= Sum of p terms
⟹a=p2{2A+(p−1)D}
⟹2ap={2A+(p−1)D} ...(i)
b= Sum of q terms
⟹b=q2{2A+(q−1)D}
⟹2bq={2A+(q−1)D} ...(ii)
and, c= Sum of r terms
⟹c=r2{2A+(r−1)D}
⟹2cr={2A+(r−1)D} ...(iii)
Multiplying equation (i), (ii) and (iii) by (q−r),(r−p) and (p−q) respectively and adding, we get
2ap(q−r)+2bq(r−p)+2cr(p−q)
={2A+(p−1)D}(q−r)+{2A+(q−1)D}(r−p)+{2A+(r−1)D}(p−q)
⟹a=p2{2A+(p−1)D}
⟹2ap={2A+(p−1)D} ...(i)
b= Sum of q terms
⟹b=q2{2A+(q−1)D}
⟹2bq={2A+(q−1)D} ...(ii)
and, c= Sum of r terms
⟹c=r2{2A+(r−1)D}
⟹2cr={2A+(r−1)D} ...(iii)
Multiplying equation (i), (ii) and (iii) by (q−r),(r−p) and (p−q) respectively and adding, we get
2ap(q−r)+2bq(r−p)+2cr(p−q)
={2A+(p−1)D}(q−r)+{2A+(q−1)D}(r−p)+{2A+(r−1)D}(p−q)
=2A(q−r+r−p+p−q)+D{(p−1)(q−1)(r−p)+(r−1)(p−q)
=2A×0+D×0=0
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