Question

solve this question for me please.
need it urgently (with steps)

Answer 1

let A be the first term and D be the common difference of the given AP. then,
a=sum of p terms
a=p/2{2A+(p-1)D}
2a/p={2A+(p-1)D}..........................1
b=sum of q terms
b=q/2{2A+(q-1)D}
2b/q={2A+(q-1)D}..........................2
and c=sum of r terns
c=r/2{2A+(r-1)D}
2c/r={2A+(r-1)D}..............................3
Multyplying eq1, eq2, eq3 by (q-r),(r-p). ,(p-q) respectively and adding them we get,
2a/p(q-r)+2b/q(r-p)+2c/r(p-q)={2A+(p-1)D}(q-r)+{2A+(q-1)D}(r-p)}+{2A+(r-1)D}(p-q)
only solving RHS side we get,
=2A(q-r)+(p-1)(q-r)D+2A(r-p)+(q-1)(r-p)D+2A(p-q)+(r-1)(p-q)D
=2A(q-r+r-p+p-q)+D{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}
=2A×0+D×0
=0
2a/p(q-r)+2b/q(r-p)+2c/r(p-q)=0
by taking 2 as a common quantity and transposing it to RHS side we get,
a/p(q-r)+b/q(r-p)+c/r(p-q)=0
HENCE PROVED