Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5
Misc 6
Misc 7 Important
Misc 8
Misc 9
Misc 10 Important
Misc 11 (i) Important
Misc 11 (ii)
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 7 Deleted for CBSE Board 2025 Exams
Question 8 Deleted for CBSE Board 2025 Exams You are here
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 14 Deleted for CBSE Board 2025 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
Question 8 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