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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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Showing 481-510 of 998 questions

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481 Easy
481 of 998
Application +4 | -1
If the first term of an AP is 3 and the sum of its 25 terms is equal to the sum of its next 15 terms, then the common difference of this AP is
View answer and explanation
Correct option: A
16\frac{1}{6}
Option A (16\frac{1}{6}) is correct because it is the common difference derived from the given conditions about the sum of terms in the arithmetic progression.
482 Easy
482 of 998
Analyzing +4 | -1
If Hn=1+12+13+…+1n  ,{H_n} = 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n}\;, then the value of Sn=1+32+53+…+2n−1n{S_n} = 1 + \frac{3}{2} + \frac{5}{3} + \ldots + \frac{{2n - 1}}{n} is
View answer and explanation
Correct option: B
2n−Hn2n - {H_n}
Option B (2n−Hn2n - {H_n}) is correct because the series can be rewritten and simplified using the definition of HnH_n.
483 Easy
483 of 998
Application +4 | -1
Let a1,a2,a3,…a_1, a_2, a_3, \ldots be terms of an A.P. If a1+a2+…+apa1+a2+…+aq=p2q2,p≠q\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+\ldots+a_q}=\frac{p^2}{q^2}, p \neq q, then a6a21\frac{a_6}{a_{21}} equals
View answer and explanation
Correct option: D
1141\frac{11}{41}
Option D (11/41) is correct because the ratio a6/a21 is calculated using the given condition and properties of A.P.
484 Medium
484 of 998
Application +4 | -1
If x,1,zx,1,z are in A.P. and x,2,zx,2,z are in G.P., then x,4,zx,4,z are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because x,4,zx, 4, z satisfy the condition for being in Harmonic Progression, which is 4=2xzx+z4 = \frac{{2xz}}{{x + z}}.
485 Easy
485 of 998
Application +4 | -1
If HH is the harmonic mean between PP and Q{\cal Q}, then the value of HP+HQ\frac{H}{P} + \frac{H}{{\cal Q}} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the harmonic mean formula simplifies to 2 when substituting and simplifying the given expression.
486 Medium
486 of 998
Application +4 | -1
If mm is the root of the equation (1−ab)x2−(a2+b2)x−(1+ab)=0(1-a b) x^2-\left(a^2+b^2\right) x-(1+a b)=0 and mm harmonic means are inserted between aa and bb, then the difference between the last and the first of the means equals
View answer and explanation
Correct option: C
ab(b−a)a b(b-a)
Option C (ab(b−a)ab(b-a)) is correct because it is derived by finding the first and last Harmonic Means and then calculating their difference using the given equation.
487 Easy
487 of 998
Application +4 | -1
Let a1,  a2,  a3.......{a_1},\,\,{a_2},\,\,{a_3}....... be an A.P. If a1+a2+...+a10a1+a2+...+ap=100p2,p≠10,\frac{{{a_1} + {a_2} + ... + {a_{10}}}}{{{a_1} + {a_2} + ... + {a_p}}} = \frac{{100}}{{{p^2}}},p \ne 10, then a11a10\frac{{{a_{11}}}}{{{a_{10}}}} is equal to
View answer and explanation
Correct option: C
2119\frac{{21}}{{19}}
Option C (2119\frac{{21}}{{19}}) is correct because it is the ratio of a11 to a10, derived from the given condition on the sum of the arithmetic progression.
488 Easy
488 of 998
Application +4 | -1
If x,2x+2,3x+3x, 2 x+2,3 x+3 are in G.P., then the fourth term is
View answer and explanation
Correct option: D
-13.5
Option D (-13.5) is correct because after finding the possible values of x, the common ratio is calculated and used to find the fourth term of the G.P.
489 Easy
489 of 998
Application +4 | -1
If first term of a GP is aa, third term is bb and (n+1)(n+1) th term is cc. The (2n+1)(2 n+1) th term of a GP is
View answer and explanation
Correct option: D
c2a\frac{c^2}{a}
Option D (c2a\frac{c^2}{a}) is correct because using the properties of a GP, we can express the (2n+1)(2n+1)th term in terms of a, b, and c.
490 Easy
490 of 998
Application +4 | -1
If log⁡10{98+x2−12x+36}=2,{\log _{10}}\left\{ {98 + \sqrt {{x^2} - 12x + 36} } \right\} = 2, then x=x =
View answer and explanation
Correct option: D
4,  84,\;8
Option D (4, 8) is correct because both values satisfy the given logarithmic equation after considering the absolute value.
491 Easy
491 of 998
Application +4 | -1
If A1,A2;G1,G2{A_1},{A_2};{G_1},{G_2} and H1,H2{H_1},{H_2} be two AM’s, GM’s and HM’s between two quantities, then the value of G1G2H1+H2\frac{{{G_1}{G_2}}}{{{H_1} + {H_2}}}
View answer and explanation
Correct option: A
A1+A2H1+H2\frac{{{A_1} + {A_2}}}{{{H_1} + {H_2}}}
Option A (A1+A2H1+H2\frac{{{A_1} + {A_2}}}{{{H_1} + {H_2}}}) is correct because, by defining the AM, GM, and HM between two quantities and using their relationships, the expression simplifies to the given form.
492 Medium
492 of 998
Application +4 | -1
The lengths of the sides of a triangle are log⁡1012\log _{10} 12, log⁡1075\log _{10} 75 and log⁡10n\log _{10} n, where n∈Nn \in N. If a and bb are the least and greatest values of nn respectively, the value of b−ab-a is divisible by
View answer and explanation
Correct option: C
223
Option C (223) is correct because the triangle inequality gives the range of possible values for nn, and the difference between the largest and smallest integer values of nn is divisible by 223.
493 Easy
493 of 998
Application +4 | -1
If the 10th term of an A.P is 120\frac{1}{{20}} and its 20th term is 110\frac{1}{{10}}, then the sum of its first 200 terms is
View answer and explanation
Correct option: D
100  12100\,\,\frac{1}{2}
Option D (100 12100\,\frac{1}{2}) is correct because the sum of the first 200 terms is calculated using the A.P. formula with the derived values of 'a' and 'd'.
494 Easy
494 of 998
Application +4 | -1
If a,b,c,da,b,c,d are in H.P., then ab+bc+cd=ab + bc + cd =
View answer and explanation
Correct option: C
3ad3ad
Option C (3ad3ad) is correct because, given that a, b, c, d are in H.P., the expression ab + bc + cd simplifies to 3ad.
495 Medium
495 of 998
Application +4 | -1
Sum of nn terms of the following series 13+33+53+73+...{1^3} + {3^3} + {5^3} + {7^3} + ... is
View answer and explanation
Correct option: A
n2(2n2−1){n^2}\left( {2{n^2} - 1} \right)
Option A (n2(2n2−1){n^2}\left( {2{n^2} - 1} \right)) is correct because it represents the sum of the given series 13+33+53+73+...{1^3} + {3^3} + {5^3} + {7^3} + ... after simplification.

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496 Easy
496 of 998
Understanding +4 | -1
The mean of the series a,a+nd,a+2nda, a+n d, a+2 n d is
View answer and explanation
Correct option: B
a+nda+n d
Option B (a+nda+nd) is correct because the mean of the series is calculated by summing the terms and dividing by the number of terms, resulting in a+nda+nd.
497 Medium
497 of 998
Application +4 | -1
The sum of the first 20 terms of the series 1+32+74+158+3116+…1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+\ldots is?
View answer and explanation
Correct option: D
38+121938+\frac{1}{2^{19}}
Option D (38+1219\frac{1}{2^{19}}) is correct because the sum of the series is calculated by recognizing the pattern and using the formula for the sum of a geometric series.
498 Easy
498 of 998
Application +4 | -1
The sum of the series 5.05 +1.212+029088+...∞+ 1.212 + 029088 + ...\infty is
View answer and explanation
Correct option: D
6.644746.64474
Option D (6.64474) is correct because it is the sum of the infinite GP calculated using the formula S = a / (1 - r).
499 Easy
499 of 998
Application +4 | -1
If the sum of nn terms of the series 23+43+63+...{2^3} + {4^3} + {6^3} + ... is 3528, then nn equals to
View answer and explanation
Correct option: D
6
Option D (6) is correct because it is the value of n that satisfies the given sum of the series.
500 Easy
500 of 998
Application +4 | -1
A person is to count 4500 currency notes. Let ana_n denotes 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 −2-2, then the time taken by him to count all notes is
View answer and explanation
Correct option: A
34 min34 \mathrm{~min}
Option A (34 min) is correct because it represents the total time taken to count all the notes, considering the initial constant rate and the subsequent arithmetic progression.
501 Easy
501 of 998
Application +4 | -1
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 2719\frac{{27}}{{19}} Then, the common ration of this series is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because it's the common ratio of the geometric series that satisfies both the sum and the sum of cubes conditions.
502 Easy
502 of 998
Application +4 | -1
The harmonic mean of 4,8 and 16 is
View answer and explanation
Correct option: C
6.85
Option C (6.85) is correct because it is the result of calculating the harmonic mean of the given numbers.
503 Easy
503 of 998
Application +4 | -1
Sum to nn terms of the series 1+(1+2)+(1+2+1+(1+2)+(1+2+ 3) +…+\ldots is
View answer and explanation
Correct option: D
16n(n+1)(n+2)\frac{1}{6} n(n+1)(n+2)
Option D (16n(n+1)(n+2)\frac{1}{6} n(n+1)(n+2)) is correct because it represents the sum to n terms of the given series, derived by summing the individual terms and simplifying.
504 Easy
504 of 998
Application +4 | -1
The value of

(1+a2x22  !+a4x44  !+…)2−(ax+a3x33  !+a5x55  !+…)2,  is{\left( {1 + \frac{{{a^2}{x^2}}}{{2\;!}} + \frac{{{a^4}{x^4}}}{{4\;!}} + \ldots } \right)^2} - {\left( {ax + \frac{{{a^3}{x^3}}}{{3\;!}} + \frac{{{a^5}{x^5}}}{{5\;!}} + \ldots } \right)^2},\;\mathrm{is}
View answer and explanation
Correct option: D
1
Option D (1) is correct because the expression simplifies to 1 using the exponential identities.
505 Easy
505 of 998
Application +4 | -1
The sum of the first 20 terms common between the series 3 +7+11+15+………+7+11+15+\ldots \ldots \ldots and 1+6+11+16+……1+6+11+16+\ldots \ldots, is
View answer and explanation
Correct option: D
4020
Option D (4020) is correct because it is the sum of the first 20 common terms of the given series, which form an arithmetic progression.
506 Medium
506 of 998
Application +4 | -1
Let a,b,ca,b,c are in GP and 4a,5b,4c4a,5b,4c are in AP such that a+b+c=70a + b + c = 70, then value of bb is
View answer and explanation
Correct option: D
20
Option D (20) is correct because using the properties of geometric and arithmetic progressions, along with the given sum, leads to the value of b being 20.
507 Easy
507 of 998
Analyzing +4 | -1
If a,  b,  c,  da,\;b,\;c,\;d are any four consecutive coefficients of any expanded binomial, then a+ba,b+cb,c+dc\frac{{a + b}}{a},\frac{{b + c}}{b},\frac{{c + d}}{c} are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the given expressions form a Harmonic Progression when a, b, c, and d are consecutive binomial coefficients.
508 Easy
508 of 998
Application +4 | -1
If log⁡2(5.2x+1),log⁡4(21−x+1)\log _2\left(5.2^x+1\right), \log _4\left(2^{1-x}+1\right) and 1 are in A.P, then xx equals
View answer and explanation
Correct option: B
1−log⁡251-\log _2 5
Option B (1−log⁡251-\log _2 5) is correct because by substituting y=2xy = 2^x and solving the resulting quadratic equation, we find x=1−log⁡25x = 1 - \log_2 5.
509 Easy
509 of 998
Application +4 | -1
The value of x=21/441/881/16161/32…x=2^{1 / 4} 4^{1 / 8} 8^{1 / 16} 16^{1 / 32} \ldots is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the infinite product can be expressed as 2 raised to the power of an infinite sum, which converges to 1.
510 Medium
510 of 998
Understanding +4 | -1
The AM, GM and HM of two numbers are x,yx, y and zz, respectively. Then, which of the following is true?
View answer and explanation
Correct option: A
z<y<xz < y < x
Option A (z<y<xz < y < x) is correct because for any two distinct positive numbers, the Arithmetic Mean (AM) is always greater than the Geometric Mean (GM), which is always greater than the Harmonic Mean (HM).
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