Question

The sum of the first p,q and r terms of an Ap is small letter a,b and c prove that a/p(q-r) + b/q(r-p) + c/r(p-q) = 0

Answer 1

Answer:

hence proved

Step-by-step explanation:

sp = p/2 ( 2a +(p-1)d)

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

2a/p = (2a+(p-1)d) _ 1

similarly

2b/q=(2a +(q-1)d)_ 2

2r/c=(2a+(r-1)d)_3

multiplying 1,2and 3by (q-r),(r-p) and (p-q) respectively and adding them

= 2a+(p-1)d)(q-r) + (2a +(q-1)d)(r-p) + (2a+(r-1)d)(p-q)

= 2a(q-r) + (p-1)d(q-r) +2a(r-p) +(q-1)d(r-p) +2a(p-q) + (r-1)d(p-q)

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

= 2a(0) + d[pq -pr- q +r + qr -pq-r+p+pr -p -qr +q]

=  0 + 0

= 0