Permutations and Combinations

Question

The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that (q – r) a + (r – p) b + (p – q)c = 0

Solution

Let A be the first term and D be the c.d. of an A.P.
 $straight t subscript straight p space equals straight a space rightwards double arrow space straight A space plus space left parenthesis straight p minus 1 right parenthesis straight D space equals space straight a$ ...(i)
$space space straight t subscript straight q space equals space straight b rightwards double arrow space straight A space plus space left parenthesis straight q minus 1 right parenthesis straight D space equals space straight b$ ...(ii)
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L.H.S. = (q - r) a + (r - p)b + (p - q)c = (q - r) [A + (p - 1)D]
 + (r - p) [A + (q - 1)D + (p - q) [A + (r - 1)D]
 (By using values of a, b, c from (i), (ii), (iii) respectively]
 = A(q - r) + (q - r) (p - 1)D + A (r - p) + (r - p) (q - 1)D + A(p - q) + (p - q) (r - 1)D
 = A(q - r + r - p + p - q) + D[Pq - q - rp + r + rq - r - pq + p + pr - p - qr + q]
 = A(0) + D(0) = 0 = R.H.S.

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Permutations And Combinations

The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that (q – r) a + (r – p) b + (p – q)c = 0

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