Series Questions

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

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 »

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

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 »

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

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 »

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

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 »

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

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 »

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

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 »

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is

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 »

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

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 »