Skip to main content
Back to SEQUENCE OF SERIES

Your Progress

0 / 998 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 961-990 of 998 questions

Preparing practice and saved progress…

961 Easy
961 of 998
Analyzing +4 | -1
If ∑r=1nTr=n(n+1)(n+2)(n+3)8\sum\limits_{{r=1}}^n T_r=\frac{n(n+1)(n+2)(n+3)}{8}, then lim⁡n→∞∑r=1n1Tr=\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{r=1}}^n \frac{1}{T_r}=
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (1/2) is correct because the limit of the sum of the reciprocals of TrT_r as n approaches infinity is 1/2.
962 Easy
962 of 998
Application +4 | -1
If log⁡e2.log⁡x27=log⁡108.log⁡e10,{\log _e}2.{\log _x}27 = {\log _{10}}8.{\log _e}10,then x=x =
View answer and explanation
Correct option: B
3
Option B (3) is correct because by simplifying the given equation using logarithm properties, we find that x = 3.
963 Easy
963 of 998
Application +4 | -1
If a1,  a2,…,  an+1{a_1},\;{a_2}, \ldots ,\;{a_{n + 1}} are in AP, then 1a1a2+1a2a3+...+1anan+1\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ... + \frac{1}{{{a_n}{a_{n + 1}}}} is
View answer and explanation
Correct option: D
na1an+1\frac{n}{{{a_1}{a_{n + 1}}}}
Option D is correct because the given expression can be simplified using the properties of arithmetic progressions, resulting in the formula n/(a₁aₙ₊₁).
964 Easy
964 of 998
Application +4 | -1
The 9th 9^{\text {th }} term of the series 27+9+525+367+⋯27+9+5 \frac{2}{5}+3 \frac{6}{7}+\cdots will be
View answer and explanation
Correct option: A
110171 \frac{10}{17}
Option A (110171 \frac{10}{17}) is correct because the series can be rewritten as 27/1,27/3,27/5,...27/1, 27/3, 27/5,..., and the 9th term is 27/(2∗9−1)=27/17=1101727/(2*9-1) = 27/17 = 1 \frac{10}{17}.
965 Easy
965 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.
966 Easy
966 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.
967 Easy
967 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.
968 Easy
968 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.
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 Easy
971 of 998
Application +4 | -1
If x=1998x=1998 !, then value of the expression 1log⁡2x+1log⁡3x\frac{1}{\log _{2} x}+\frac{1}{\log _{3} x} +……+1log⁡1998x+\ldots \ldots+\frac{1}{\log _{1998} \mathrm{x}} equals-
View answer and explanation
Correct option: C
1
Option C (1) is correct because using the change of base formula and properties of logarithms simplifies the expression to 1.
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
Analyzing +4 | -1
If 1a+1c+1a−b+1c−b=0\frac{1}{a} + \frac{1}{c} + \frac{1}{{a - b}} + \frac{1}{{c - b}} = 0 and b≠a+c,b \ne a + c, then a,b,ca,b,c are in
View answer and explanation
Correct option: B
H.P.
Option B (H.P.) is correct because the given equation simplifies to the condition for a,b,ca, b, c to be in Harmonic Progression.
975 Easy
975 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.
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 Easy
978 of 998
Application +4 | -1
If the function f:R→Rf: R \rightarrow R satisfies f(x+y)=f(x)+f(y),∀ x,y∈Rf(x+y)=f(x)+f(y), \forall \, x, y \in \mathrm{R} and f(1)=7f(1)=7, then ∑r=1nf(r)\sum\limits_{{r=1}}^n f(r)
View answer and explanation
Correct option: D
7n(n+1)2\frac{7 n(n+1)}{2}
Option D (7n(n+1)2\frac{{7 n(n+1)}}{{2}}) is correct because the function f(r) equals 7r, and the sum of 7r from r=1 to n is 7n(n+1)2\frac{{7 n(n+1)}}{{2}}.
979 Easy
979 of 998
Analyzing +4 | -1
The sum of the series 1x+1+2x2+1+22x4+1+....+2100x2100+1 \frac{1}{{x + 1}} + \frac{2}{{{x^2} + 1}} + \frac{{{2^2}}}{{{x^4} + 1}} + .... + \frac{{{2^{100}}}}{{{x^{{2^{100}}}} + 1}}\,, when x=2x = 2 is
View answer and explanation
Correct option: D
1−21014101−11 - \frac{{{2^{101}}}}{{{4^{101}} - 1}}
The correct answer is D (1−21014101−11 - \frac{{{2^{101}}}}{{{{4^{101}} - 1}}}) because it is the result of telescoping the series and substituting x = 2.
980 Easy
980 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.
981 Easy
981 of 998
Application +4 | -1
If 13+23+33+…+n3=20251^3+2^3+3^3+\ldots+n^3=2025, then n=n=
View answer and explanation
Correct option: B
9
Option B (9) is correct because it satisfies the equation derived from the sum of cubes formula, leading to the correct value of n.
982 Easy
982 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.
983 Easy
983 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).
984 Easy
984 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.
985 Easy
985 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.
986 Easy
986 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.
987 Easy
987 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.
988 Easy
988 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.
989 Easy
989 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.
990 Easy
990 of 998
Analyzing +4 | -1
The value of log⁡3e−log⁡9e+log⁡27e−log⁡81e+…ad.inf.,{\log _3}e - {\log _9}e + {\log _{27}}e - {\log _{81}}e + \ldots \mathrm{ad}.\mathrm{inf}., is
View answer and explanation
Correct option: B
log⁡32{\log _3}2
Option B (log⁡32{\log _3}2) is correct because the series simplifies to a logarithmic expression using properties of logarithms and the series expansion of ln(2).
Revision notes and key takeaways

Review SEQUENCE OF SERIES in your textbook or class notes, then use this set to check your understanding.

Showing 961-990 of 998 questions

Your practice

0 / 998 answered · 998 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.