Misc 15

Transcript

Misc 15 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 Here we have small ‘a’ in the equation, so we use capital ‘A’ for first term We know that, An = A + (n – 1) D where An is the nth term of A.P. n is the number of terms A is the first term, D is the common difference It is given that pth term of an AP is a i.e. Ap = a Putting n = p A + (p – 1) D = a a = A + (p – 1)D qth term of an AP is b i.e. Aq = b Putting n = q A + (q – 1)D = b b = A + (q – 1)D rth term of an AP is c i.e. Ar = c Putting n = r A + (r – 1)D = c c = A + (r – 1)D Now we need to show that (q – r)a + (r – p)b + (p – q)c = 0 We have a = A + (p – 1)D Multiplying by (q – r) a (q – r) = [A + (p – 1)D] (q – r) a (q – r) = [A (q – r) + (q – r) (p – 1)D] And b = [A + (q – 1)D] Multiplying by (r – p) b (r – p) = [A + (q – 1)D] (r – p) b (r – p) = [A (r – p) + (q – r) (q – 1)D] Similarly, c = A + (r – 1)D multiply this by (p – q) c (p – q) = [A + (r – 1)D] (p – q) c (p – q) = [A (p – q) + (p – q) (r – 1)D] We have to prove a (q – r) + b (r – p) + c (p – q) = 0 Taking L.H.S a (q – r) + b (r – p) + c (p – q) From (1), (2) & (3) = "[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) + A (r – p) + A (p – q) + (q – r) (p – 1)D" "+ (r – p) (q – 1)D + (p – q) (r – 1)D" = A[q – r + r – p + p – q] + D[(q – r) (p – 1) + (r – p) (q – 1) + (p – q) (r – 1)] = A[p – p + r – r + q – q] + D[q(p – 1) – r(p – 1) + r(q – 1) – p(q – 1) + p(r – 1) – q(r – 1)] = A[0 + 0 + 0] + D[qp – q – rp + r + rq – r – pq + p + pr – p – qr + q] = 0 + D[qp – pq + rq – qr – rp + pr + q – q + r – r + p – p ] = 0 + D[0] = 0 = R.H.S Hence proved