Solution : Since, x, y and z are in AP \(\therefore\) 2y = x + z Also, \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are in AP \(\therefore\) 2\(tan^{-1}y\) = \(tan^{-1}x\) + \(tan^{-1}z\) \(\implies\) \(tan^{-1}({2y\over {1 – y^2}})\) = \(tan^{-1}({x + z\over {1 – xz}})\) \(\implies\) \(x + z\over {1 – y^2}\) = \(x + z\over {1 – […]
Solution : 0.7 + 0.77 + 0.777 + …… + upto 20 terms = \(7\over 10\) + \(77\over 10^2\) + \(777\over 10^3\) + ….. + upto 20 terms = 7[ \(1\over 10\) + \(11\over 10^2\) + \(111\over 10^3\) + ….. + upto 20 terms ] = \(7\over 9\)[ \(9\over 10\) + \(99\over 100\) + \(999\over
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is Read More »
Solution : Let a, ar, \(ar^2\) are in GP (r > 1) According to the question, a, 2ar, \(ar^2\) in AP. \(\implies\) 4ar = a + \(ar^2\) \(\implies\) \(r^2\) – 4r + 1 = 0 \(\implies\) r = \(2 \pm \sqrt{3}\) Hence, r = \(2 + \sqrt{3}\) [ \(\because\) AP is increasing] Similar Questions
Solution : \(K(10)^9\) = \((10)^9\) + \(2(11)^1(10)^8\) + \(3(11)^2(10)^7\) + …… + \(10(11)^9\) K = 1 + 2\(({11\over 10})\) + 3\(({11\over 10})^2\) + ….. + 10\(({11\over 10})^9\) ……(i) \(({11\over 10})\)K = 1\(({11\over 10})\) + 2\(({11\over 10})^2\) + 3\(({11\over 10})^3\) + ….. + 10\(({11\over 10})^{10}\) …..(ii) On subtracting equation (ii) from (i),
Solution : Let the probability of getting a head be p and not getting a head be q. Since, head appears first time in an even throw 2 or 4 or 6. \(\therefore\) \(2\over 5\) = qp + \(q^3\)p + \(q^5\)p + …… \(\implies\) \(2\over 5\) = \(qp\over {1 – q^2}\) Since q = 1-
Solution : Let \(A_1\), \(A_2\), \(A_3\) be the events of match winning in first, second and third matches respectively and whose probabilities are P(\(A_1\)) = P(\(A_2\)) = P(\(A_2\)) = \(1\over 2\) \(\therefore\) Required Probability = P(\(A_1\))P(\(A_2’\))P(\(A_3\)) + P(\(A_1’\))P(\(A_2\))P(\(A_3\)) = \(({1\over 2})^3\) + \(({1\over 2})^3\) = \(1\over 8\) + \(1\over 8\) = \(1\over 4\) Similar Questions
Solution : Probability of getting success, p = \(1\over 6\) and probability of failure, q = \(5\over 6\) Now, we must get 2 sixes in seven throws, so probability is \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\) and the probability that 8th throw is \(1\over 6\). \(\therefore\) Required Probability = \(^7C_2\)\(({1\over 6})^2\)\(({5\over 6})^5\)\(1\over 6\) = \(^7C_2\times 5^5\over {6^8}\) Similar
Question : If A and B are two mutually exclusive events, then (a) P(A) < P(B’) (b) P(A) > P(B’) (c) P(A) < P(B) (d) None of these Solution : Since, A and B are two mutually exclusive events. \(\therefore\) A \(\cap\) B = \(\phi\) \(\implies\) Either A \(\subseteq\) B’ or B \(\subseteq\) A’
If A and B are two mutually exclusive events, then Read More »
Solution : The total number of ways in which numbers can be choosed = 25*25 = 625 The number of ways in which either players can choose same numbers = 25 \(\therefore\) Probability that they win a prize = \(25\over 625\) = \(1\over 25\) Thus, the probability that they will not win a prize in
Solution : Since, total number of students = 100 and number of boys = 70 \(\therefore\) number of girls = (100 – 70) = 30 Now, the total marks of 100 students = 100*72 = 7200 And total marks of 70 boys = 70*75 = 5250 Total marks of 30 girls = 7250 – 5250
