If pth, qth and rth terms of an A.P. are a, b, c respectively, then show that (i) a(q-r)+b(r-p)+c(p-q)=0

If pth, qth and rth terms of an A.P. are a, b, c respectively, then sh

1.	If pth, qth and rth terms of an A.P. are a, b, c respectively, then show that (i) a(q-r)+b(r-p)+c(p-q)=0
Answer» Let x be the first term and d be the common difference of the given AP. Then, `T_(p) = x + (p-1)d, T_(q) = x +(q-1)d "and" T_(r) = x + (r-1)d.` `therefore x + (p-1)d =a " "…(i)` `x + (p-1)d = b " "...(ii)` `x +(r-1)d = c " "... (iii)` On multiplying (i) (q-r), (ii) by (r-p) and (iii) by (p-q), and 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 xx 0) + (d xx 0) = 0.` Hence, a(q-r) +b(r-p) +c (p-q) = 0.