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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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541 Medium
541 of 564
Application +4 | -1
Let x1,x2,…∈(0,π)x_1, x_2, \ldots \in(0, \pi) denote the of values of xx satisfying the equation 27(1+∣cos⁡x∣+cos⁡2x+∣cos⁡x∣3+… upto ∞)=9327^{\left(1+|\cos x|+\cos ^2 x+|\cos x|^3+\ldots \text { upto } \infty\right)}=9^3, find the value of 1π(x1+x2+…)\frac{1}{\pi}\left(x_1+x_2+\ldots\right)
View answer and explanation
Correct option: A
1
The correct answer is None of these because A50A^{50} is [10251]\left[\begin{array}{ll}1 & 0 \\ 25 & 1\end{array}\right], which is not among the provided options.
542 Medium
542 of 564
Application +4 | -1
If the ppth term of an AP be q  and  qq\;\mathrm{and}\;qth term be pp, then its rrth term will be
View answer and explanation
Correct option: B
p+q−rp + q - r
Option B (p+q−rp + q - r) is correct because it is derived from solving the system of equations formed by the given conditions of the arithmetic progression.
543 Medium
543 of 564
Application +4 | -1
If A.M., G.M. and H.M. of first and last terms of the series 100,101,102,….n−1,n100,101,102, \ldots . \mathrm{n}-1, \mathrm{n} are the terms of the series it self then the value of n\mathrm{n} is (100<n≤500)(100<\mathrm{n} \leq 500)
View answer and explanation
Correct option: C
400
Option C (400) is correct because it satisfies the conditions for the A.M., G.M., and H.M. to be integers within the given range.
544 Medium
544 of 564
Application +4 | -1
The value of nn for which 704+12(704)+14(704)+…n704+\frac{1}{2}(704)+\frac{1}{4}(704)+\ldots \quad n \quad terms =1984−12(1984)+14(1984)−…n=1984-\frac{1}{2}(1984)+\frac{1}{4}(1984)-\ldots n upto terms, is
View answer and explanation
Correct option: A
5
Option A (5) is correct because solving the equation derived from the given geometric series results in n=5 when n is odd.
545 Medium
545 of 564
Application +4 | -1
If the roots of the equation x3−12x2+39x−28=0x^3-12 x^2+39 x-28=0 are in AP, then the common difference will be
View answer and explanation
Correct option: C
±3\pm 3
Option C (±3\pm 3) is correct because it represents the common difference of the AP whose terms are the roots of the given cubic equation, found using Vieta's formulas.
546 Medium
546 of 564
Application +4 | -1
The sum of three numbers in GP is 56 . If we subtract 1,7,211,7,21 from these numbers in that order, we obtain an APA P. Then, three numbers are
View answer and explanation
Correct option: A
8,16,328,16,32
Option A (8, 16, 32) is correct because these numbers are in GP and satisfy the condition of forming an AP after subtraction.
547 Medium
547 of 564
Analyzing +4 | -1
If SkS_k denotes the sum of first kk terms of a G.P. Then, Sn,S2n−Sn,S3n−SnS_n, S_{2 n}-S_n, S_{3 n}-S_n are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because the terms SnS_n, S2n−SnS_{2n} - S_n, and S3n−S2nS_{3n} - S_{2n} form a geometric progression.
548 Medium
548 of 564
Application +4 | -1
Find PP if 3+14(3+P)+142(3+2P)+…∞=83+\frac{1}{4}(3+P)+\frac{1}{4^2}(3+2 P)+\ldots \infty=8
View answer and explanation
Correct option: A
9
Option A (9) is correct because it is the value of P that satisfies the given infinite series equation after simplification and solving for P.
549 Medium
549 of 564
Application +4 | -1
If 7 th and 13 th terms of an AP are 34 and 64 respectively, then its 18 th term is
View answer and explanation
Correct option: C
89
Option C (89) is correct because it is the 18th term of the AP, calculated using the given 7th and 13th terms to find the common difference and first term.
550 Medium
550 of 564
Application +4 | -1
The sum of the integers from 1 to 100 which are divisible by exactly one of 2 and 5 is
View answer and explanation
Correct option: D
3050
Option D (3050) is correct because it is the sum of integers from 1 to 100 divisible by 2 or 5, but not both.
551 Medium
551 of 564
Analyzing +4 | -1
Consider the sequence of numbers 121, 12321, 1234321, … Each term in the sequence is
View answer and explanation
Correct option: B
Square of an odd number
Option B (Square of an odd number) is correct because each term in the sequence is the square of a number consisting of all 1s, which is an odd number.
552 Medium
552 of 564
Application +4 | -1
IQ of a person is given by the formula IQ=MACA×100\mathrm{IQ}=\frac{\mathrm{MA}}{\mathrm{CA}} \times 100 where, MA is mental age and CA is chronological age.If 80≤IQ≤14080 \leq \mathrm{IQ} \leq 140 for a group of 12 years children, then the range of their mental age is
View answer and explanation
Correct option: C
9.6≤MA≤16.89.6 \leq \mathrm{MA} \leq 16.8
Option C (9.6≤MA≤16.89.6 \leq \mathrm{{MA}} \leq 16.8) is correct because it represents the range of mental age (MA) calculated from the given IQ range and chronological age.
553 Medium
553 of 564
Application +4 | -1
Let f:R→Rf:R \to R be such that for all x∈Rx \in R (21+x+21−x),f(x)\left( {{2^{1 + x}} + {2^{1 - x}}} \right),f\left( x \right) and (3x+3−x)\left( {{3^x} + {3^{ - x}}} \right) are in A.P, then the minimum value of f(x)f\left( x \right) is
View answer and explanation
Correct option: B
3
Option B (3) is correct because the minimum value of f(x)f(x) is found by using the AM-GM inequality.
554 Medium
554 of 564
Application +4 | -1
If the product of three terms of G.P. is 512. If 8 added to first and 6 added to second term, so that number may be in A.P., then the numbers are
View answer and explanation
Correct option: B
4,8,164,8,16
Option B (4,8,16) is correct because the product is 512 and adding 8 to the first term and 6 to the second results in an arithmetic progression.
555 Medium
555 of 564
Analyzing +4 | -1
If a2,b2,c2{a^2},{b^2},{c^2} are in AP, then which of the following is also an AP?
View answer and explanation
Correct option: C
cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C
Option C (cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C) is correct because if a2,b2,c2{a^2},{b^2},{c^2} are in AP, then cot⁡A,cot⁡B,cot⁡C\cot A,\cot B,\cot C are also in AP, which can be derived using trigonometric identities and the sine rule.

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556 Medium
556 of 564
Application +4 | -1
The rational number which is equal to the number 2.357‾2 . \overline{357} with recurring decimal is
View answer and explanation
Correct option: C
2355999\frac{2355}{999}
Option C (2355999\frac{2355}{999}) is correct because it represents the given recurring decimal as a fraction.
557 Medium
557 of 564
Analyzing +4 | -1
f tn=14(n+2)(n+3)t_n=\frac{1}{4}(n+2)(n+3) for n=1,2,3,…n=1,2,3, \ldots, then 1t1+1t2+1t3+….+1t2020=\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\ldots .+\frac{1}{t_{2020}}=
View answer and explanation
Correct option: D
80806069\frac{8080}{6069}
Option D (80806069\frac{8080}{6069}) is correct because it is the result of telescoping the series and simplifying the expression.
558 Medium
558 of 564
Understanding +4 | -1
The sum of infinite terms of the series 1(1+a)(2+a)+1(2+a)(3+a)+1(3+a)(4+a)+...+to  ∞\frac{1}{{\left( {1 + a} \right)\left( {2 + a} \right)}} + \frac{1}{{\left( {2 + a} \right)\left( {3 + a} \right)}} + \frac{1}{{\left( {3 + a} \right)\left( {4 + a} \right)}} + ... + \mathrm{to}\;\infty, where aa is a constant, is
View answer and explanation
Correct option: A
11+a\frac{1}{{1 + a}}
Option A (11+a\frac{1}{{1 + a}}) is correct because the series telescopes, and the limit of the partial sum as n approaches infinity is 11+a\frac{1}{{1 + a}}.
559 Medium
559 of 564
Application +4 | -1
If for x∈(0, π2),  log⁡10sin⁡x+log⁡10cos⁡x=−1x \in \left( {0,\,\frac{\pi }{2}} \right),\,\,{\log _{10}}\sin x + {\log _{10}}\cos x = - 1 and log⁡10(sin⁡x+cos⁡x)=12(log⁡10n−1),  n>0{\log _{10}}\left( {\sin x + \cos x} \right) = \frac{1}{2}\left( {{{\log }_{10}}n - 1} \right),\,\,n > 0, then the value of nn is equal to:
View answer and explanation
Correct option: C
12
Option C (12) is correct because solving the logarithmic equations and using trigonometric identities leads to n = 12.
560 Medium
560 of 564
Application +4 | -1
21/4⋅41/8⋅81/16……∞2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \ldots \ldots \infty is equal to -
View answer and explanation
Correct option: B
2
Option B (2) is correct because the infinite product can be expressed as a power of 2, where the exponent is an infinite sum that converges to 1.
561 Medium
561 of 564
Application +4 | -1
The sum to n\mathrm{n} terms of the series [11+12+14+21+22+24+31+32+34+…..+n1+n2+n4]\left[\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+\ldots . .+\frac{\mathrm{n}}{1+\mathrm{n}^2+\mathrm{n}^4}\right] is
View answer and explanation
Correct option: A
n(n+1)2(n2+n+1)\frac{n(n+1)}{2\left(n^2+n+1\right)}
Option A n(n+1)2(n2+n+1)\frac{{n(n+1)}}{{2(n^2+n+1)}} is correct because the sum telescopes after rewriting the general term.
562 Medium
562 of 564
Application +4 | -1
Ar,r=1,2,3,…,nA_r, r=1,2,3, \ldots, n are nn points on the parabola y2=4xy^2=4 x in the first quadrant. If Ar=(xr,yr)A_r=\left(x_r, y_r\right), where x1,x2,x3,…,xnx_1, x_2, x_3, \ldots, x_n are in GP and x1=1x_1=1, x2=2x_2=2, then yny_n is equal to
View answer and explanation
Correct option: A
2n+122^{\frac{n+1}{2}}
Option A (2n+122^{\frac{n+1}{2}}) is correct because the y-coordinates form a GP with a common ratio of 2\sqrt{2}, and the nth term can be calculated.
563 Medium
563 of 564
Application +4 | -1
The set of real values of xx for which log⁡0.2x+2x≤1\log _{0.2} \frac{x+2}{x} \leq 1 is
View answer and explanation
Correct option: A
(−∞,−52]∪(0,∞)\left(-\infty,-\frac{5}{2}\right] \cup(0, \infty)
Option A ((−∞,−52-\infty,-\frac{5}{2}] ∪\cup(0, ∞\infty)) is correct because it represents the set of real values of x that satisfy the given logarithmic inequality.
564 Medium
564 of 564
Application +4 | -1
If a1,a2,…..,ana_1, a_2, \ldots . ., a_n are in A.P. with common difference dd, then the sum of the series, sec⁡a1sec⁡a2+sec⁡a2sec⁡a3+…..+sec⁡an−1sec⁡an=k(tan⁡an−tan⁡a1)\sec a_1 \sec a_2+\sec a_2 \sec a_3+\ldots . .+\sec a_{n-1} \sec a_n=k\left(\tan a_n-\tan a_1\right) where k=k=
View answer and explanation
Correct option: C
cosec⁡d\operatorname{cosec} d
Option C is correct because the sum telescopes, and the common difference in the A.P. allows us to express the terms in a way that simplifies the sum.
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