If p^th ,qth and rth terms of an ap are a,b,c respectively then show t

1.	If p^th ,qth and rth terms of an ap are a,b,c respectively then show that (q-r)+b(r-p)+c(p-q)=0
Answer» Given that the pth, qth and rth terms of an AP be a, b, c respectively.Suppose, x be the first term and d be the common difference of the given arithmetic progression. Therefore,Tp = x+(p-1)d, Tq =x+(q-1)d and Tr =x+(r-1)dNow, Tp\xa0= a\xa0{tex}\\Rightarrow{/tex}\xa0x+(p-1)d = a ....(i)Tq\xa0= b\xa0{tex}\\Rightarrow{/tex}\xa0x+(q-1)d = b....(ii)Tr\xa0= c\xa0{tex}\\Rightarrow{/tex} x+(r-1)d = c....(iii)On multiplying (i) by (q-r), (ii) by (r-p) and (iii) by (p-q), we get,(q-r)[x+(p-1)d]=a(q-r)....(iv)(r-p)[x+(q-1)d]=b(r-p)....(v)(p-q)[x+(r-1)d]=c(p-q)....(vi)Now adding we get,a(q - r) + b(r - p) + c(p - q) = x{(q - r) + (r - p) + (p - q)} + d{(p - 1)(q - r)+ (q - 1)(r - p)+ (r - 1)(p - q)}= (x\xa0{tex}\\times{/tex}\xa00) + (d {tex}\\times{/tex}\xa00) = 0Therefore, a(q - r) + b(r - p) + c(p - q) = 0.Hence\xa0proved.

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