If pth, qth, rth terms are a,b,c.then prove that p (b-c)+q (c-a)+r (a-b)=0
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Heya friend .
let x be the first term and d be the common difference of the given AP .
then ,
tp=x+(p-1)d
tq=x+(q-1)d=b
tr=x+(r-1)d
•°•x+(p-1)d=a
x+(q-1)d=b
x+(r-1)d=c
on multiplication (1) by (2)by (r-p) and (3)by (p-1)
and adding ,we get.
a(q-r)+b(r-p)+c(p-q)=x*[(q-r)+(r-p)+(p-q)]+d*{(q-r)+(r-p)+(r-p)+(p-q)}+d*{p-1)(q-1)(r-p)+(r-1)(p-q)
=(x×0)+(d×0)=0
Hence ,a(q-r)+b(r-p)+c(p-q)=0
Hope it helps you.
@Rajukumar
let x be the first term and d be the common difference of the given AP .
then ,
tp=x+(p-1)d
tq=x+(q-1)d=b
tr=x+(r-1)d
•°•x+(p-1)d=a
x+(q-1)d=b
x+(r-1)d=c
on multiplication (1) by (2)by (r-p) and (3)by (p-1)
and adding ,we get.
a(q-r)+b(r-p)+c(p-q)=x*[(q-r)+(r-p)+(p-q)]+d*{(q-r)+(r-p)+(r-p)+(p-q)}+d*{p-1)(q-1)(r-p)+(r-1)(p-q)
=(x×0)+(d×0)=0
Hence ,a(q-r)+b(r-p)+c(p-q)=0
Hope it helps you.
@Rajukumar
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