Series Questions

Solution : Let a be the first term and r the common ratio of the given G.P. Then, \(a_4\) = 54  and  \(a_9\) = 13122 \(\implies\)  \(ar^3\) = 54   and  \(ar^8\)  =  13122 \(\implies\)  \(ar^8\over ar^3\) = \(13122\over 54\)  \(\implies\)  \(r^5\) = 245  \(\implies\)  r = 3 Putting r = 3 in \(ar^3\) = 54, […]

Solution : By using method of differences, The difference between the successive terms are 7 – 3 = 4, 14 – 7 = 7, 24 – 14 = 10, …. Clearly,these differences are in AP. Let \(T_n\) be the nth term and \(S_n\) denote the sum to n terms of the given series Then, \(S_n\)

Solution : By using method of differences, The difference between the successive terms are 15 – 3 = 12, 35 – 15 = 20, 63 – 35 = 28, …. Clearly,these differences are in AP. Let \(T_n\) be the nth term and \(S_n\) denote the sum to n terms of the given series Then, \(S_n\)

Solution : Let \(S_n\) = 1 + 2 + 3 + ….. + n = \(\sum_{k=1}^{n} k\) Clearly, it is an arithmetic series with first term a = 1, common difference d = 1 and last term l = n. \(\therefore\) \(S_n\) = \(n\over 2\) (1 + n) = \(n(n+1)\over 2\) Similar Questions Let \(a_n\)

Solution : Given, \(a_2 + a_4 + a_6 + …… + a_{200}\) = \(\alpha\)      ………(i) and \(a_1 + a_3 + a_5 + ….. + a_{199}\) = \(\beta\)           ………(ii) On subtracting equation (ii) from equation (i), we get (\(a_2 – a_1\)) + (\(a_4 – a_3\)) + ……… + (\(a_{200} –

Solution : Let the time taken to save Rs 11040 be (n + 3) months. for first three months, he saves Rs 200 each month. In (n + 3) months, 3 \(\times\) 200 + \(n\over 2\) { 2(240) + (n – 1) \(\times\) 40 } = 11040 \(\implies\)  600 + \(n\over 2\) {40(12+ n –

Solution : Let a be the first term and d (d \(\ne\) 0) be the common difference of the given AP, then \(T_{100}\) = a + (100 – 1)d = a + 99d \(T_{50}\) = a + (50 – 1)d = a + 49d \(T_{150}\) = a + (150 – 1)d = a + 149d

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

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