Solution : As given \(\bar{x}\) = \(x_1 + x_2 + …. + x_n\over n\) If the mean of the series \(x_i\) + 2i, i = 1, 2, ….., n be \(\bar{X}\), then \(\bar{X}\) = \((x_1+2) + (x_2+2.2) + (x_3+2.3) + …. + (x_n + 2.n)\over n\) = \(x_1 + x_2 + …. + x_n\over n\) …
Solution : By using method of differences, The \(n^{th}\) term is (2n-1)(2n+1)(2n+3) \(T_n\) = (2n-1)(2n+1)(2n+3) \(T_n\) = \(1\over 8\)(2n-1)(2n+1)(2n+3){(2n+5) – (2n-3)} = \(1\over 8\)(\(V_n\) – \(V_{n-1}\)) \(S_n\) = \({\sum}_{r=1}^{n} T_n\) = \(1\over 8\)(\(V_n\) – \(V_0\)) \(\therefore\) \(S_n\) = \((2n-1)(2n+1)(2n+3)(2n+5)\over 8\) + \(15\over 8\) = \(n(2n^3 + 8n^2 + 7n – 2)\) Similar Questions Find the …
Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + …… Read More »
Solution : \(\because\) \(T_n\) = \(S_n – S_{n-1}\) = \({\sum}_{r=1}^{n} T_r\) – \({\sum}_{r=1}^{n – 1} T_r\) = \(n(n+1)(n+2)(n+3)\over 8\) – \((n-1)(n)(n+1)(n+2)\over 8\) = \(n(n+1)(n+2)\over 8\)[(n+3) – (n-1)] = \(n(n+1)(n+2)\over 8\)(4) \(T_n\) = \(n(n+1)(n+2)\over 2\) \(\implies\) \(1\over T_n\) = \(2\over n(n+1)(n+2)\) = \((n+2)-n\over n(n+1)(n+2)\) = \(1\over n(n+1)\) – \(1\over (n+1)(n+2)\) Let \(V_n\) = \(1\over n(n+1)\) \(\therefore\) …
Solution : Clearly P(A) = Power set of A = set of all subsets of A = {\(\phi\), {x}, {y}, {x,y}} Similar Questions Let A and B be two sets containing 2 elements and 4 elements, respectively. The number of subsets A\(\times\)B having 3 or more elements is If aN = {ax : x \(\in\) …
Solution : 6N = {6, 12, 18, 24, 30, …..} 8N = {8, 16, 24, 32, ….} \(\therefore\) 6N \(\cap\) 8N = {24, 48, …..} = 24N Similar Questions If A = {x,y}, then the power set of A is If A = {2, 4} and B = {3, 4, 5} then (A \(\cap\) B) …
If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Read More »
Solution : A = [x: x \(\in\) R,-1 < x < 1] B = [x : x \(\in\) R, x – 1 \(\le\) -1 or x – 1 \(\ge\) 1] [x: x \(\in\) R, x \(\le\) 0 or x \(\ge\) 2] \(\therefore\) A \(\cup\) B = R – D where D = [x : x …
Solution : \(x^2\) = 16 \(\implies\) x = \(\pm\)4 2x = 6 \(\implies\) x = 3 There is no value of x which satisfies both the above equations. Thus, A = \(\phi\) Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then the …
The set A = [x : x \(\in\) R, \(x^2\) = 16 and 2x = 6] equal Read More »
Solution : R = {(2, 4), (4, 3), (6, 2), (8, 1)} \(R^{-1}\) = {(4, 2), (3, 4), (2, 6), (1, 8)} Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then the set 6N \(\cap\) 8N is equal to Let A = …
Solution : (A \(\cap\) B) = {4} and (A \(\cup\) B) = {2, 3, 4, 5} \(\therefore\) (A \(\cap\) B) \(\times\) (A \(\cup\) B) = {(4, 2), (4, 3), (4, 4), (4, 5)} Similar Questions If A = {x,y}, then the power set of A is If aN = {ax : x \(\in\) N}, then …
If A = {2, 4} and B = {3, 4, 5} then (A \(\cap\) B) \(\times\) (A \(\cup\) B) Read More »
Solution : Given P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2 \(\therefore\) P(A) + P(B) + P(C) = 1 then events A, B, C are exhaustive. If P(E) = Probability of introducing a new product, then as given P(E|A) = 0.7, P(E|B) = 0.6 and P(E|C) = 0.5 = 0.5 \(\times\) 0.7 + …
