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IIT JEE Previous Years Questions: Part 1 (in Hindi)
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Solution of IIT JEE previous year questions.

Shivam Gupta
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sir why does current remains same in series combination and voltage varies and the vice versa occurs in Parallel combination ?
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  1. JEE PREVIOUS YEAR SOLUTION SEQUENCE AND SERIES


  2. Question Let V denotes the sum of the first r terms of an AP, whose first term is r and common difference is [IIT 2007] (2t-1). Then, the sum + + + is ()n(n 1)(3n2+n+2) (d) (2n3-2n + 3) (a) n(n + 1)3n2-n+1) 12 12 (c)n(2n2-n1) Solution: We have, K = 5Zr + (r-1)(2-1)) | Sn = n [2a + (n-1)d] 2 1 [ m(n1)2 n(n +1)(2n1) n(n +1) n(n + 1)(3n2 + n + 2) = 12


  3. Question: 2 [IIT 2007] Let V denotes the sum of the first r terms of an AP, whose first term is r and common difference is (2t-1). Then, T, = Vr+1-K-2 is always (a) an odd number (c) a prime number (b) an even number (d) a composite number Solution: We have Tr = (3-1)(r + 1) , which is composite number. Composite Number A composite number is a positive integer that can be formed by multiplying together two smaller positive integers T, 2(r + 1)3-(r + 1)2 + r + 1)--(2p3-y2 + r)-2


  4. Question: 3 [IIT 2007] Let V denotes the sum of the first r terms of an AP, whose first term is r and common difference is(2t-1). Then, Tr = Vr+1-K-2 and Qr-Tr+1-7, . Which of the following correct statement? (a) 01, Q2, Q3,.. are in AP with common difference 5 (b) Q1, 02, Q3, .. are in AP with common difference 6 (c) Q1, Q2, Q3, are in AP with common difference 11 Solution: We have, Qr 3 1)2 +2(r 1) 1 322r -1) Qr = 6r + 5 Q1 = 11, Q2-17, Q3-23 So Qr is the rh term of an AP with common difference 6


  5. uestion: 4If the sum of first n terms of an AP is cn, then the sum of squares of these n terms is IIT 2009] n(4n2-1)c2 n(4n2+1)c2 (c) n2-1)2 n(4n2+1)c2 Solution: Let S cn2 Sn = 4c2 n(n+1)(2n + 1) nc2(4n2-1 3


  6. Question: 5 Let a1, a2 be a harmonic progression with a1 = 5 and a20 = 25 IIT 2012] The least positive integer n for which an < 0, is (a) 22 (b) 23 (c) 24 (d) 25 1 95-4n + 4 an Solution: It is given that a1,a2,a3,.. are in HP, therefore .. are in AP. Let d be the common difference of the AP -- 19 25 99 4n a1 a2 a3 Now an < 0 99-4n 19 25 =+ (n-1)d and = + 19d a1 a20 at a20 a1 4n > 99 25 5 4 19 25 11 4(n - 1) an 5 19 x 25