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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
JEE Mains and Advanced Mathematics with Shivam Sir Be maths expert with us - Aspire Study Youtube Channel NIMCET PYQs

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thanks mam ..you are superb .. agar aap jara aur hindi mai samjhye to mere jaise aur students brilliant ban jaaye .. English ke aap phir se hindi mai uss topic ko hindi mai bhi bataye
I really 🎩 😁 👕👍Great! 👖 🎩 😁 👕👍Great! 👖 🎩 😁 👕👍Great! 👖 🎩 😁 👕👍Great! 👖 🎩 😁 👕👍Great! 👖 🎩 😁 👕👍Great! 👖 🎩 😁 👕👍Great! 👖 thank you sir
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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