Maths Questions

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 »

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), …

If \((10)^9\) + \(2(11)^1(10)^8\) + \(3(11)^2(10)^7\) + …… + \(10(11)^9\) = \(K(10)^9\), then k is equal to Read More »