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, …

If the 4th and 9th terms of a G.P. be 54 and 13122 respectively, find the G.P. Read More »

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, \(S_n\) …

Find the sum of n terms of the series 3 + 7 + 14 + 24 + 37 + ……. Read More »

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, \(S_n\) …

Find the sum to n terms of the series : 3 + 15 + 35 + 63 + ….. Read More »

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 \(a_n\) …

Prove that the sum of first n natural numbers is \(n(n+1)\over 2\) Read More »

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} – …

Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is Read More »

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 – …

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after Read More »

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 …

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is Read More »

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 – …

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then Read More »

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 …

Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is Read More »