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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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Showing 961-990 of 998 questions
961 Medium
961 of 998
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.
962 Easy
962 of 998
Application +4 | -1
If sum of the first 2nn terms of an AP series 2, 5, 8,… is equal to the sum of the first nn terms of the AP series 57, 59, 61,…, then nn equals
View answer and explanation
Correct option: C
11
Option C (11) is correct because it is the value of n that equates the sum of the first 2n terms of the first AP series to the sum of the first n terms of the second AP series.
963 Medium
963 of 998
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.
964 Easy
964 of 998
Application +4 | -1
If tan⁡(π9), x, tan⁡(7π18)\tan \left( {\frac{\pi }{9}} \right),\,x,\,\tan \left( {\frac{{7\pi }}{{18}}} \right) are in arithmetic progression and tan⁡(π9), y, tan⁡(5π18)\tan \left( {\frac{\pi }{9}} \right),\,y,\,\tan \left( {\frac{{5\pi }}{{18}}} \right) are also in arithmetic progression, then ∣x−2y∣\left| {x - 2y} \right| is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because using the arithmetic progression property and trigonometric identities, the expression |x - 2y| simplifies to 0.
965 Medium
965 of 998
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.
966 Easy
966 of 998
Application +4 | -1
The sum of the series
11+2+12+3+13+4+…\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots

upto 15 terms is
View answer and explanation
Correct option: C
3
Option D (12\frac{1}{2}n) is correct because by using the property of binomial coefficients and summing the series in both forward and reverse order, we can derive the relationship between t_n and S_n.
967 Easy
967 of 998
Application +4 | -1
A person is to count 4500 currency notes. Let ana_n denote the number of notes he counts in the nnth minute. If a1=a2=…=a10=150a_1=a_2=\ldots=a_{10}=150 and a10,a11,…a_{10}, a_{11}, \ldots are in AP with common difference - 2, then the time taken by him to count all notes is
View answer and explanation
Correct option: D
34 minutes
Option D (34 minutes) is correct because after calculating the notes counted in the first 10 minutes, we solve a quadratic equation to find the additional time required, and choose the valid solution.
968 Medium
968 of 998
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.
969 Easy
969 of 998
Application +4 | -1
If a,b,ca,b,c are in A.P., b−a,c−bb - a,c - b and a are in G.P., then a:b:ca:b:c is
View answer and explanation
Correct option: A
1  :2  :31\;:2\;:3
Option A is correct because using the given conditions that a, b, c are in A.P. and b-a, c-b, a are in G.P., we can derive the ratio a:b:c as 1:2:3.
970 Easy
970 of 998
Application +4 | -1
If 9a2+4b2=18ab,9{a^2} + 4{b^2} = 18ab, then log⁡(3a+2b)=\log \left( {3a + 2b} \right) =
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the correct simplification of the equation leads to a different logarithmic expression than the given options.
971 Medium
971 of 998
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.
972 Easy
972 of 998
Application +4 | -1
If x,y,zx,y,z are in H.P., then log⁡(x+z)+log⁡(x−2y+z)\log \left( {x + z} \right) + \log \left( {x - 2y + z} \right) is equal to
View answer and explanation
Correct option: B
2log⁡(x−z)2\log \left( {x - z} \right)
Option B (2log⁡(x−z)2\log \left( {x - z} \right)) is correct because by using the properties of harmonic progression and logarithms, we can simplify the expression to 2log(x-z).
973 Easy
973 of 998
Application +4 | -1
0.14189189189… can be expressed as a rational number
View answer and explanation
Correct option: D
21148\frac{{21}}{{148}}
Option D (21148\frac{{21}}{{148}}) is correct because the repeating decimal 0.14189189189... can be converted to a fraction by expressing it as a sum of fractions and simplifying.
974 Easy
974 of 998
Application +4 | -1
Let a1,a2,a3,…a_1, a_2, a_3, \ldots be a harmonic progression with a1=5a_1=5 and a20=25a_{20}=25. The least positive integer nn for which an<0a_n<0, is
View answer and explanation
Correct option: D
25
Option D (25) is correct because it is the smallest integer n for which an<0a_n < 0, derived by analyzing the harmonic progression and its corresponding arithmetic progression.
975 Medium
975 of 998
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.
976 Easy
976 of 998
Application +4 | -1
Let f(x)f\left( x \right) be a polynomial function of second degree. If f(1)=f(−1)f\left( 1 \right) = f\left( { - 1} \right) and a,b,ca,b,c are in AP, then f′(a),  f′(b)f'\left( a \right),\;f'\left( b \right) and f′(c)f'\left( c \right) are in
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the derivative of a quadratic function is a linear function, and if the original values are in AP, their corresponding values in the linear function will also be in AP.
977 Easy
977 of 998
Application +4 | -1
If 3rd ,6th &3^{\text {rd }}, 6^{\text {th }} \& last term of HP is 13,15,3203\frac{1}{3}, \frac{1}{5}, \frac{3}{203}. Then the number of terms is
View answer and explanation
Correct option: D
100
Option D (100) is correct because by converting the HP to AP, we find the common difference and first term, then solve for the number of terms.
978 Medium
978 of 998
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.
979 Easy
979 of 998
Application +4 | -1
If the sum of an infinite GP and the sum of square of its term is 3, then the common ratio of the first series is
View answer and explanation
Correct option: D
32\frac{3}{2}
Option B (12\frac{1}{2}) is correct because it is the common ratio of the infinite GP that satisfies both the sum of the series and the sum of the squares of its terms being equal to 3.
980 Easy
980 of 998
Understanding +4 | -1
Consider the following statements.

I. A sequence is called an arithmetic progression if the difference of a term and the previous term is always same.

II. Arithmetic Mean (A.M.) AA of any two numbers aa and bb is given by 12(a+b)\frac{1}{2}(a+b) such that a,A,ba, A, b are in A.P. The arithmetic mean for any nn positive numbers a1,a2,a3,….,ana_1, a_2, a_3, \ldots ., a_n is given by

A.M. =a1+a2+….+ann\text {A.M. }=\frac{a_1+a_2+\ldots .+a_n}{n}

Choose the correct option.
View answer and explanation
Correct option: B
Both are true
Option B (Both are true) is correct because both statements accurately describe the definition of an arithmetic progression and its arithmetic mean.
981 Easy
981 of 998
Application +4 | -1
If log⁡8x=2.5{\log _8}x = 2.5 and log⁡8x=2.5{\log _8}x = 2.5 then x=x =
View answer and explanation
Correct option: A
y3/2{y^{3/2}}
Option A (y3/2{y^{3/2}}) is correct because by converting the logarithmic equations to exponential form and solving for x in terms of y, we find that x = y(3/2).
982 Medium
982 of 998
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.
983 Easy
983 of 998
Analyzing +4 | -1
Let S=85+1615+...+128218+1,S = \frac{8}{5} + \frac{{16}}{{15}} + ... + \frac{{128}}{{{2^{18}} + 1}},, then
View answer and explanation
Correct option: A
S=1088545S = \frac{{1088}}{{545}}
Option A is correct because the given sum can be expressed as a telescoping series, which simplifies to the given value.
984 Easy
984 of 998
Application +4 | -1
The first term of a GP is 7, the last term is 448 and sum of all term is 889, then the common ratio is
View answer and explanation
Correct option: D
2
Option D (2) is correct because it is the common ratio that satisfies the given conditions for the geometric progression.
985 Easy
985 of 998
Application +4 | -1
The sum of first 9 terms of the series 131+13+231+3+13+23+331+3+5+…\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\ldots is
View answer and explanation
Correct option: C
96
Option C (96) is correct because it is the sum of the first 9 terms of the given series, calculated by simplifying the general term and then summing the series.
986 Easy
986 of 998
Application +4 | -1
If a,b,ca,b,c be in arithmetic progression, then the value of (a+2b−c)(2b+c−a)(a+2b+c)\left( {a + 2b - c} \right)\left( {2b + c - a} \right)\left( {a + 2b + c} \right) is
View answer and explanation
Correct option: A
16abc16abc
Option A (16abc16abc) is correct because when a,b,ca, b, c are in arithmetic progression, 2b=a+c2b = a + c, which simplifies the expression to 16abc16abc.
987 Easy
987 of 998
Application +4 | -1
Let a1,a2,…,a9a_1, a_2, \ldots, a_9 be in G.P. with a1<0a_1 < 0 such that a1+a2=4,a3+a4=16a_1+a_2=4, a_3+a_4=16. If ∑i=19ai=4λ\sum\limits_{i=1}^9 a_i=4 \lambda, then λ\lambda is equal to
View answer and explanation
Correct option: C
−171-171
Option C is correct because by finding the common ratio and the first term of the G.P., the sum of the first 9 terms can be calculated, leading to λ=−171\lambda = -171.
988 Medium
988 of 998
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}}.
989 Medium
989 of 998
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.
990 Easy
990 of 998
Application +4 | -1
Consider an infinite geometric series with first term a and common ratio rr. If its sum is 4 and the second term is 34\frac{3}{4}, then
View answer and explanation
Correct option: D
a=3,  r=14a = 3,\;r = \frac{1}{4}
Option D (a=3,  r=14a = 3,\;r = \frac{1}{4}) is correct because these values satisfy both the sum of the infinite geometric series and the value of the second term.
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