In the print window choose “Save as PDF”. The file prints without a web address or date line.
Mere Teacher · Previous Year Questions
JEE Main Mathematics · Sequence & Series · Set 9 (Q241-270)
Scan or visit mereteacher.com for free PYQ practice, timed CBT tests, step-by-step solutions and BroBot AI. Answers and solutions are free with a sign-up.
241.
Let an be nth term of the series 5 + 8 + 14 + 23 + 35 + 50+.......and Sn=∑k=1nak. Then S30 – S40 is equal to
(1)11310
(2)11260
(3)11290
(4)11280
242.
Let A1, A2, A3, ...... be an increasing geometric progression of positive real numbers. If A1A3A5A7=12961 and A2 + A4 =367, then, the A6+A8+A10value of A6 + A8 + A10 is equal to
(1)33
(2)37
(3)43
(4)47
243.
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777,....., is
(1)817 (179 – 10–20)
(2)97 (99 – 10–20)
(3)817 (179 + 10–20)
(4)97 (99 + 10–20)
244.
If the sum of the first ten terms of the series (153)2+(252)2+(351)2+42+(454)2+...... is 516m, then m is equal to :
(1)101
(2)100
(3)99
(4)102
245.
If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is
(1)21
(2)22
(3)23
(4)24
246.
nth term of the series 1+73+725+737+… is
(1)7n2n−1
(2)7n−12n−1
(3)7n−12n+1
(4)7n2n+1
247.
Consider an A.P.: a1,a2,…,an; a1>0. If a2−a1=4−3, an=41a1, and ∑i=1nai=2525, then ∑i=117ai is equal to
(1)136
(2)476
(3)238
(4)952
248.
The value of ∑k=1∞(−1)k+1(k!k(k+1)) is
(1)2/e
(2)1/e
(3)e/2
(4)e
249.
The sum ∑k=120k2k1 is equal to:
(1)2−2173
(2)1−22011
(3)2−21911
(4)2−22021
250.
If n=2020!, then log2n1+log3n1+log4n1+…+log2020n1=
(1)2020
(2)0
(3)2020
(4)1
251.
Let a1,a2,a3,… be a G.P. such that a1<0, a1+a2=4 and a3+a4=16. If ∑i=19ai=4λ, then λ is equal to:
(1)−513
(2)−3511
(3)−171
(4)171
252.
The parabolas ax2+2bx+cy=0 and dx2+2ex+fy=0 intersect on the line y=1. If a,b,c,d,e,f are positive real numbers and a,b,c are in G.P., then
(1)d,e,f are in A.P
(2)ad,be,cf are in G.P
(3)ad,be,cf are in A.P
(4)d,e,f are in G.P
253.
Let a1,a2,a3,… be in an A.P. such that ∑k=112a2k−1=−572a1,a1=0. If ∑k=1nak=0, then n is:
(1)18
(2)17
(3)11
(4)10
254.
Let an be the nth term of a G.P. of positive terms. If ∑n=1100a2n+1=200 and ∑n=1100a2n=100, then ∑n=1200an
(1)175
(2)150
(3)300
(4)225
255.
Let a,b,c>1, a2,b2 and c2 be in A.P., and log3b, logea and logec be in G.P. If the sum of first 20 terms of an A.P., whose first term is 3a+4b+c and the common difference is 10a−8b+c is −444. Then abc is equal to
(1)343
(2)216
(3)8343
(4)8125
256.
For three positive integers p,q,r, xpq2=yqr=zp2r and r=pq+1 such that 3,3logyx,3logzy,7logxz are in A.P. with common difference 21, then r−p−q is equal to
(1)2
(2)6
(3)12
(4)−6
257.
If 0<θ,φ<2π, x=∑n=0∞cos2nθ, y=∑n=0∞sin2nφ and z=∑n=0∞cos2nθ⋅sin2nφ then
(1)xyz=4
(2)xy−z=(x+y)z
(3)xy+yz+zx=z
(4)xy+z=(x+y)z
258.
Let y=f(x) represent a parabola with focus (−21,0) and directrix y=−21. Then S={x∈R:tan−1(f(x))+sin−1(f(x)+1)=2π}
(1)contains exactly two elements
(2)Contains exactly one element
(3)Is an infinite set
(4)Is an empty set
259.
3+5+7+… to n terms is
(1)n2
(2)n(n+2)
(3)n(n−2)
(4)(n+1)2
260.
If log32,log3(2x−5),log3(2x−27) are in an arithmetic progression, then the value of x is equal to
261.
If (20)19+ 2(21)(20) 18 + 3(21) (20) 17 +... +20(21) 19 = k(20) 19, then k is equal to_____.
262.
The sum to 20 terms of the series 2.22 - 32 + 2.42 - 52 + 2. 62 -............is equal to___________.
263.
Let a1, a2, a3, ....., a11 be real numbers satisfying a1 = 15, 27 – 2a2 > 0 and ak = 2ak–1 – ak–2 for k = 3, 4, ...., 11. If 11a12+a22+....+a121 = 90, then the value of 11a1+a2+...+a11 is equal to
264.
If a1(> 0), a2, a3, a4, a5 are in a G.P., a2 + a4 = 2a3 + 1 and 3a2 + a3 = 2a4, then a2 + a4 + 2a5 is equal to
265.
Let Sk, k = 1, 2,...., 100, denote the sum of the infinite geometric series whose first term isk!k–1 and the common ratio is k1. Then the value of 100!1002+ ∑k=1100(k2−3k+1)Sk is
266.
Let (ak) and (bk), k∈N, be two G.P.s with common ratio r1 and r2 respectively such that a1=b1=4 and r1<r2. Let ck=ak+bk, k∈N. If c2=5 and c3=413 then ∑k=1∞ck−(12a6+8b4) is equal to _____
267.
Suppose a1,a2,2,a3,a4 be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is 249, then a4 is equal to______________
268.
Let ∑n=0∞(n!)((2n)!)n3((2n)!)+(2n−1)(n!)=ae+eb+c Where a,b,c∈Z and e=∑n=0∞n!1 Then a2−b+c is equal to __________
269.
Let A1,A2,A3 be the three AP. with the same common difference d having their first terms as A,A+1,A+2, respectively. Let a,b,c be 7th, 9th, 17th terms of A1,A2,A3, respectively such that a2bc71717111+70=0
270.
If ∑r=125(r4+r2+1r)=qp, where p and q are positive integers such that gcd(p,q)=1, then p+q is equal to ____.