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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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Showing 631-660 of 998 questions

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631 Easy
631 of 998
Application +4 | -1
If log⁡2,log⁡(2x−1)\log 2, \log \left(2^x-1\right) and log⁡(2x+3)\log \left(2^x+3\right) are in APA P, then the value of xx is given by
View answer and explanation
Correct option: B
log⁡25\log _2 5
Option B (log⁡25\log _2 5) is correct because it is the solution obtained by setting up and solving the equation derived from the arithmetic progression property.
632 Easy
632 of 998
Application +4 | -1
The sum of the series 13×7+17×11+111×15+...\frac{1}{{3 \times 7}} + \frac{1}{{7 \times 11}} + \frac{1}{{11 \times 15}} + ... is
View answer and explanation
Correct option: D
112\frac{1}{{12}}
Option D (112\frac{1}{{12}}) is correct because it is the sum of the infinite series, calculated by expressing each term as a difference and observing the telescoping nature of the series.
633 Easy
633 of 998
Application +4 | -1
Suppose a1=45,a2=41a_1=45, a_2=41 and ak=2ak−1−ak−2 ∀ k≥3a_k=2 a_{k-1}-a_{k-2} \, \forall \, k \geq 3, then, value of

S=112[a12−a22+a32−a42+…+a112−a122]S=\frac{1}{12}\left[a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{11}^2-a_{12}^2\right] is
View answer and explanation
Correct option: D
92
Option D (92) is correct because the given recurrence relation implies an arithmetic progression, and the sum is calculated correctly using the properties of an A.P.
634 Easy
634 of 998
Understanding +4 | -1
If SS is the sum of an infinite GP, the first term aa, then the common ratio rr is given by
View answer and explanation
Correct option: B
S−aS\frac{{S - a}}{S}
Option B (S−aS\frac{{S - a}}{S}) is correct because the common ratio can be derived from the formula for the sum of an infinite geometric progression.
635 Easy
635 of 998
Application +4 | -1
If the sum of second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of its first 50 terms is
View answer and explanation
Correct option: D
126(350−1)\frac{1}{26}\left(3^{50}-1\right)
Option B (2+32+\sqrt{3}) is correct because it is the common ratio that satisfies the given conditions of the increasing G.P. and the resulting A.P.
636 Easy
636 of 998
Application +4 | -1
The sum of all natural numbers less than 200 that are divisible neither by 3 nor by 5 , is
View answer and explanation
Correct option: B
10732
Option B (10732) is correct because it represents the sum of natural numbers less than 200 that are not divisible by 3 or 5, calculated using inclusion-exclusion principle.
637 Medium
637 of 998
Application +4 | -1
Let b1,b2,……….b_1, b_2, \ldots \ldots \ldots .. , be a geometric sequence such that b1+b2=1b_1+b_2=1 and ∑k=1∞bk=2\sum\limits_{{k=1}}^{\infty} b_k=2. Given that b2<0b_2<0, then the value of b1b_1 is:
View answer and explanation
Correct option: C
2+22+\sqrt{2}
Option C (2+22+\sqrt{2}) is correct because it's derived by solving the equations formed by the given conditions of the geometric sequence and considering the constraint b2 < 0.
638 Medium
638 of 998
Application +4 | -1
If the sum of nn terms of an AP is 3n2−n3 n^2-n and its common difference is 6 , then its first term is
View answer and explanation
Correct option: A
2
Option A (2) is correct because by finding the general term of the AP using the sum of n terms, we can determine the first term.
639 Easy
639 of 998
Application +4 | -1
If a,b,ca,b,c are in G.P., then log⁡aλ,log⁡bλ,log⁡cλ{\log _a}\lambda ,{\log _b}\lambda ,{\log _c}\lambda are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the logarithms form a Harmonic Progression since a, b, c are in GP.
640 Easy
640 of 998
Application +4 | -1
If the ratio of sums to nn terms of two A.P.'s is (5n+7):(3n+2)(5 n+7):(3 n+2), then the ratio of their 17th terms is
View answer and explanation
Correct option: B
172:101172: 101
Option A (172: 101) is correct because it is the ratio of the 17th terms of the two arithmetic progressions, calculated using the given ratio of their sums.
641 Easy
641 of 998
Application +4 | -1
If nn geometric means be inserted between aa and bb, then the nnth geometric mean will be
View answer and explanation
Correct option: C
a(ba)nn+1a{\left( {\frac{b}{a}} \right)^{\frac{n}{{n + 1}}}}
Option C (a(ba)nn+1a{\left( {\frac{b}{a}} \right)^{\frac{n}{{n + 1}}}}) is correct because it represents the nth geometric mean when n geometric means are inserted between a and b.
642 Easy
642 of 998
Application +4 | -1
If x=−2,x = - 2, then the value of log⁡4(x24)−2log⁡44(x4),{\log _4}\left( {\frac{{{x^2}}}{4}} \right) - 2{\log _4}4\left( {{x^4}} \right), is
View answer and explanation
Correct option: C
−6- 6
Option C is correct because by substituting x = -2 into the logarithmic expression and simplifying using logarithmic properties, the value is found to be -6.
643 Easy
643 of 998
Application +4 | -1
The sum to infinity of the GPG P −54,516,−564,…\frac{-5}{4}, \frac{5}{16}, \frac{-5}{64}, \ldots, is
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because the sum to infinity of the given GP is calculated using the formula S = a/(1-r) where a = -5/4 and r = -1/4.
644 Easy
644 of 998
Application +4 | -1
The first and last term of an A.P. are aa and ll respectively. If SS be the sum of all the terms of the A.P., then the common difference is
View answer and explanation
Correct option: A
l2−a22S−(l+a)\frac{l^2-a^2}{2 S-(l+a)}
Option A is correct because the common difference is derived by using the formulas for the last term and the sum of an A.P.
645 Easy
645 of 998
Application +4 | -1
If 9 AM’s and HM’s are inserted between the 2 and 3 and if the harmonic mean HH is corresponding to arithmetic mean A,  then  A+6HA,\;\mathrm{then}\;A + \frac{6}{H} is equal to
View answer and explanation
Correct option: C
5
Option C (5) is correct because by inserting 9 arithmetic and harmonic means between 2 and 3, and using the properties of AP and HP, we find that A+6H=5A + \frac{6}{H} = 5.

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646 Medium
646 of 998
Understanding +4 | -1
If the arithmetic mean and geometric mean of aa and bb are AA and GG respectively, then the value of A−GA-G will be
View answer and explanation
Correct option: C
[a−b2]2\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2
Option C (a−b22\frac{{\sqrt{{a}} - \sqrt{{b}}}}{{\sqrt{{2}}}}^2) is correct because it is the simplified form of the difference between the arithmetic mean and geometric mean of a and b.
647 Easy
647 of 998
Analyzing +4 | -1
The value of 2−12+3−224+52−762+17−12216+…+…+ad.  inf.,\frac{{\sqrt 2 - 1}}{{\sqrt 2 }} + \frac{{3 - 2\sqrt 2 }}{4} + \frac{{5\sqrt 2 - 7}}{{6\sqrt 2 }} + \frac{{17 - 12\sqrt 2 }}{{16}} + \ldots + \ldots + \mathrm{ad}.\mathrm{\;inf}., is
View answer and explanation
Correct option: B
log⁡e2{\log _e}\sqrt 2
Option B is correct because the given series can be rewritten as a logarithmic series, and its sum converges to log⁡e2{\log _e}\sqrt 2.
648 Hard
648 of 998
Analyzing +4 | -1
The sum of the products of the numbers ±1,±2,…±n\pm 1, \pm 2, \ldots \pm n, taken two at a time is
View answer and explanation
Correct option: C
−n(n+1)(2n+1)6\frac{{ - n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}
Option C (−n(n+1)(2n+1)6\frac{{ - n(n + 1)(2n + 1)}}{6}) is correct because the sum of the products of the numbers is derived from the square of the sum of the numbers, accounting for the squares of individual numbers.
649 Easy
649 of 998
Application +4 | -1
The sum to infinity of the series 1+2(1−1n)+3(1−1n)2+…..1+2\left(1-\frac{1}{n}\right)+3\left(1-\frac{1}{n}\right)^2+….. , is
View answer and explanation
Correct option: A
n2n^2
Option A (n2n^2) is correct because it is the sum to infinity of the given series, derived by using the arithmetico-geometric progression formula.
650 Medium
650 of 998
Application +4 | -1
If a,ba, b and cc are in AP\mathrm{AP}, then (a+2b−c)(2b+c−a)(a+2 b-c)(2 b+c-a) (c+a−b)(c+a-b) equals
View answer and explanation
Correct option: D
4abc4 a b c
Option D (4abc4abc) is correct because the expression simplifies to 4abc4abc when 2b=a+c2b = a + c is substituted, given that a, b, and c are in AP.
651 Medium
651 of 998
Application +4 | -1
Let pnp_n denote the probability of getting nn heads when a fair coin is tossed mm times. If P4,P5,P6P_4, P_5, P_6 are in Arithmetic Progression, then the sum of all possible values of mm is
View answer and explanation
Correct option: B
21
Option B (21) is correct because by setting up the arithmetic progression condition for the probabilities and solving for m, the sum of all possible values of m is found to be 21.
652 Easy
652 of 998
Application +4 | -1
Let a1,a2,a3,……,an,…a_1, a_2, a_3, \ldots \ldots, a_n, \ldots be in A.P. If a3+a7+a11+a15=72a_3+a_7+a_{11}+a_{15}=72. Then the sum of its first 17 terms is equal to
View answer and explanation
Correct option: B
306
Option B (306) is correct because the sum of the first 17 terms is calculated using the given condition and the properties of an A.P.
653 Easy
653 of 998
Application +4 | -1
If the 2nd, 5th and 9th terms of a non-constant AP are in GP, then the common ratio of this GP is
View answer and explanation
Correct option: B
43\frac{4}{3}
The correct answer is B (43\frac{4}{3}) because the common ratio is calculated using the given condition that the 2nd, 5th, and 9th terms are in GP.
654 Easy
654 of 998
Application +4 | -1
The 5th term of the sequence 109,13203,23,…  \frac{{10}}{9},\frac{1}{3}\frac{{\sqrt {20} }}{3},\frac{2}{3}, \ldots \; is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option D (25\frac{2}{5}) is correct because it is the 5th term of the given geometric progression, calculated using the formula for the nth term of a GP.
655 Easy
655 of 998
Application +4 | -1
The number of terms of the G.P. 3,  32,  34, ...3,\,\,\frac{3}{2},\,\,\frac{3}{4},\,... needed to obtain a sum of 3069512\frac{3069}{512} is:
View answer and explanation
Correct option: B
10
Option B (10) is correct because by using the geometric progression sum formula, we can find the number of terms needed to obtain a sum of 3069512\frac{3069}{512}.
656 Easy
656 of 998
Application +4 | -1
If log⁡xa−b=log⁡yb−c=log⁡zc−a,\frac{{\log x}}{{a - b}} = \frac{{\log y}}{{b - c}} = \frac{{\log z}}{{c - a}}, then xyzxyz is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because when the given logarithmic ratios are equal, it implies that xyz=1xyz = 1.
657 Easy
657 of 998
Application +4 | -1
If x=∑n=0∞(−1)ntan⁡2nθx=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta and y=∑n=0∞cos⁡2nθy=\sum_{n=0}^{\infty} \cos ^{2 n} \theta, for 0<θ<π40<\theta<\frac{\pi}{4}, then:
View answer and explanation
Correct option: B
y(1−x)=1y(1-x)=1
Option B (y(1-x)=1) is correct because it is derived by expressing x and y as infinite geometric series and then finding a relationship between them.
658 Hard
658 of 998
Application +4 | -1
If x,y,zx,y,z are three consecutive positive integers, then

12log⁡ex+12log⁡ez+12  x  z+1+13(12  x  z+1)3+…\frac{1}{2}{\log _e}x + \frac{1}{2}{\log _e}z + \frac{1}{{2\;x\;z + 1}} + \frac{1}{3}{\left( {\frac{1}{{2\;x\;z + 1}}} \right)^3} + \ldots is equal to
View answer and explanation
Correct option: B
log⁡ey{\log _e}y
Option B (log⁡ey{\log _e}y) is correct because by using the properties of logarithms and the given condition that x, y, and z are consecutive integers, the expression simplifies to log⁡ey{\log _e}y.
659 Easy
659 of 998
Analyzing +4 | -1
The value of ax−bya^x-b^y is (where x=log⁡abx=\sqrt{\log _a b} and y=log⁡ba,a>0,b>0y=\sqrt{\log _b a}, a>0, b>0 and a,b≠1)a, b \neq 1)
View answer and explanation
Correct option: C
0
Option C (0) is correct because by substituting the values of x and y, we find that a^x = b^y, thus a^x - b^y = 0.
660 Easy
660 of 998
Application +4 | -1
The value of 5log⁡577log⁡75{5^{\sqrt {{{\log }_5}7} }}{7^{\sqrt {{{\log }_7}5} }} is
View answer and explanation
Correct option: C
0
Option D (None of these) is correct because the expression simplifies to zero, which is not among the given options. The correct answer should be 0.
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