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This lesson takes up various problems in sequences series

## Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE | 6 Million Minutes Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com

U
Sir Please Explain light in seperate Course.
Sushant dubey
a year ago
kk
sir aap bohot acchha padhate hai...thank u very much
why 1st option is not possible in question1
why 1st option is not possible in question1
why 1st option is not possible in question1
why 1st option is not possible in question1
1. SOLVED EXAMPLES SEQUENCES SERIES JEE MAIN RANK BOOSTER COURSE PREPARED BY: ER. VINEET LOOMBA ITIAN IIT-JEE MENTOR

2. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME ** B.Tech. From IIT Roorkee -\$ IIT-JEE Mentor Since 2010. -^ Doubts/Feedback in Comment Section * Comment other topics you want to revise. * Follow me @ https://unacademy.com/user/vineetloomba to get unacademy updates or search me on Google * Share among your peers as SHARING is CARING!!

3. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Other Detailed Courses Made so far on Unacademv: V Strategy for JEE Main and Advanced V Sets, Relations and Functions V Trigonometry V Applications of Derivatives Limits, Continuity. Differentiability V Indefinite Integration V Definite Integration V Complex Number V Logarithmic Functions V Sequences Series V Most Important Questions in IIT-JEE Permutations Combinations Binomial Theorem V Straight Lines V Applications of Integrals V Circles (Detailed Course) V Probability MathematicS V Parabola (Detailed Course) v Inverse Trigonometry V Mathematical Reasoning V Ellipse (Detailed Course) V Hyperbola Upcoming Courses next month: Coordinate Geometry > Differential Equations Differential Equations MATHEMATI ER. VINEET LOOMBA (IIT RooRKee)

4. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 1. If first and (2n-1)th terms of an A.P. , G. P. and H.P., are equal and their nth terms are a, b, c respectively, then (A) a+c 2b (C) asbsc (B) a c b (D) ac-b20 Sol. (C). Let be the first , b, will be GP. , c, will be in HP Hence a, b, c are respectively A. M., G.M. and H.M. of a and Since A.M. 2 G.M. 2H.M., a 2b2C Again a = +p , b2 = and c-2ap. Hence ac-b2=0. and be the (2n-1) terns of an A.P., GP. and H.P., then , a, will be in AP., 2 + SUCCESS IN JEE MAIN 2019 ER. VINEET LOOMBA (IIT RooRKEE)

5. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 2. If a, b, c and d are distinct positive numbers in H.P. , then (A) a+b>c+d (C) a+d>b+ c (B) a+c> b+d (D) none of these a + C 2 b+d 2 Sol. (C). Since b is the H.M. of a and c, b (A.M.> H.M.) Again c is the H.M. of b and d ,->c (AM.> H.M.) Adding, we get 2+d Adding, we get_ >b+ C a+d>b+ C 2 SUCCESS IN JEE MAIN 2019 ER. VINEET LOOMBA (IIT RooRKEE)

6. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 3. The value of natural number a for which f(a+k) 16(2-1) where the function f satisfies fx+y)x) fy) for all natural numbers x, y and f(1) (A) 2 (C) 5 2 is equal to (D) 8 Sol.(B) It is given that fx + y) f(x) f(y) and f(1) 2.f (2) f(1 1) f1)f(1) 2.2 22 f(3) -f(1) f(2) 2.22 23f(k) 2 and f(a)2a k-1 k-1 k-1 - f(a)[2 +22+2 2(1-2") 16(2" -1) f(a). 16(2" -1) 20.2(2" -1) SUCCESS IN EE MAIN 2019 ER. VINEET LOOMBA (IIT RooRKEe)

7. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) The sum of the series (n).1 + (n-1).2 + (n-2).3 + .....+ 1.n is n(n2 +1 6 n(n-1 6 n(n +1 (n +2 6 6 Sol.(A). The rth term of the series is given by Tr (n-r+1)r sum of the series- T-(n + 1) r- r2 2m S 233 n(n +1 2 n(n+1(n+2) SUCCESS IN EE MAIN 2019 ER. VINEET LOOMBA (IIT RooRKEe)

8. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) a, +a2+a p Let a1, a2, a3,... be terms of an A.P. If 2, p q , t hen equals 21 = a, a2+ a 2 2a, (p-1d q 2a, +(q-1)d q Sol.D) 2a, (q-1)d For , p=11, q=41 4-11 21 21 SUCCESS IN EE MAIN 2019 ER. VINEET LOOMBA (IIT RooRKEe)

9. 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) The number of terms common to the two A.P's 3,7,11,. 407 and 2, 9,16, (A) 12 (C) 17 ,709 is equal to (B) 14 (D) none of these Sol (B). Let number of terms in 2 A.P's be m and n respectively. Then, 407 mth terms of 1st A.P. and 709 nth term of second A.P 407 3 (m 1)4 and 709 2+(n 1) x 7 m-102 and n-102 so, each A.P. has 102 terms Let pth term of 1st A.P. be identical to qth term o 2nd A.P. Then, 4p -17q-5 7k 1 102 and 4k 102 k 14 and k25 2 ks 14 => k = 1, 2, 3, , 14 SUCCESS IN EE MAIN 2019 ER. VINEET LOOMBA (IIT RooRKEe)

10. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ENROL for this Course (Free) Enroll 439 Recommend Lessons (Like) E Rate and Review the Course .Comments . Sharing with friends 15 68 111 ratings 21 reviews Share MATHEMATI ER. VINEET LOOMBA (IIT RooRKEE)