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 · Mixed difficulty

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 91-120 of 998 questions

Preparing practice and saved progress…

91 Easy
91 of 998
Application +4 | -1
If b−c,2b−λ,b−ab-c, 2 b-\lambda, b-a are in HPH P, then a−λ2a-\frac{\lambda}{2}, b−λ2,c−λ2b-\frac{\lambda}{2}, c-\frac{\lambda}{2} are is
View answer and explanation
Correct option: B
GP
The correct answer is GP because after the given harmonic progression is manipulated, it results in the condition for a geometric progression.
92 Easy
92 of 998
Application +4 | -1
If ax=by=cz=dw,{a^x} = {b^y} = {c^z} = {d^w}, then log⁡a(bcd){\log _a}\left( {bcd} \right) equals to
View answer and explanation
Correct option: B
x(1y+1z+1w)x\left( {\frac{1}{y} + \frac{1}{z} + \frac{1}{w}} \right)
Option B (x(1y+1z+1w)x\left( {\frac{1}{y} + \frac{1}{z} + \frac{1}{w}} \right)) is correct because by taking logarithms and manipulating the equations, log⁡a(bcd){\log _a}\left( {bcd} \right) can be expressed as x(1y+1z+1w)x\left( {\frac{1}{y} + \frac{1}{z} + \frac{1}{w}} \right).
93 Easy
93 of 998
Application +4 | -1
If <an>< {a_n} > is an arithmetic sequence, then Δ  =∣amanapmnp111∣\Delta \; = \left| {\begin{array}{cccccccccccccccccccc} {{a_m}}&{{a_n}}&{{a_p}}\\ m&n&p\\ 1&1&1 \end{array}} \right| equals
View answer and explanation
Correct option: C
0
Option C (0) is correct because the determinant is zero when the first row is a linear combination of the other rows, which is the case for an arithmetic sequence.
94 Medium
94 of 998
Application +4 | -1
If x>0x>0, then the sum of the series e−x−e−2x+e−3x−…e^{-x}-e^{-2 x}+e^{-3 x}-\ldots is
View answer and explanation
Correct option: D
11+ex\frac{1}{1+e^x}
Option D (11+ex\frac{{1}}{{1+e^x}}) is correct because it is the sum of the given infinite geometric series.
95 Easy
95 of 998
Application +4 | -1
If x,  1,  zx,\;1,\;z are in AP and x,  2,  zx,\;2,\;z are in GP, then x,  4,  zx,\;4,\;z will be in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because given the conditions for AP and GP, the resulting relationship indicates that x, 4, z are in Harmonic Progression.
96 Medium
96 of 998
Application +4 | -1
13−23+33−43+…+93=1^{3}-2^{3}+3^{3}-4^{3}+\ldots+9^{3}=
View answer and explanation
Correct option: A
425
Option A (425) is correct because the sum of the series can be calculated by grouping terms and using the formula for the sum of cubes.
97 Easy
97 of 998
Application +4 | -1
Let f(x)f(x) be a polynomial of second degree. If f(1)=f(−1)f(1)=f(-1) and a,ba, b, cc are in A.P., then f′(a),f′(b)f^{\prime}(a), f^{\prime}(b) and f′(c)f^{\prime}(c) are in
View answer and explanation
Correct option: B
A.P.
Option B (A.P.) is correct because the derivative of the quadratic function results in a linear function, and since a, b, c are in A.P., their corresponding values in the linear function are also in A.P.
98 Easy
98 of 998
Application +4 | -1
The least value of ' aa ' for which 51+x+51−x,a/2,25x+25−x5^{1+x}+5^{1-x}, a / 2,25^x+25^{-x} are three consecutive terms of an A.P., is
View answer and explanation
Correct option: C
12
Option C (12) is correct because it is the least value of 'a' that satisfies the arithmetic progression condition, obtained by using AM-GM inequality.
99 Easy
99 of 998
Application +4 | -1
The value of the sum ∑r=1n∑s=1nSrs  2r3s,whereSrs=0,  if  r≠s  andSrs=1,    if  r=s,\sum \limits_{r = 1}^n \sum \limits_{s = 1}^n {S_{rs}}\;{2^r}{3^s},\mathrm{where}{S_{rs}} = 0,\;\mathrm{if}\;r \ne s\;\mathrm{and}{S_{rs}} = 1,\;\;\mathrm{if}\;r = s, is
View answer and explanation
Correct option: B
65(6n−1)\frac{6}{5}\left( {{6^n} - 1} \right)
Option B is correct because the sum simplifies to a geometric progression with a common ratio of 6, allowing for direct calculation using the geometric series formula.
100 Easy
100 of 998
Application +4 | -1
If S=1+a+a2+……S=1+a+a^2+\ldots \ldots to ∞(a<1)\infty(a<1), then the value of aa, is
View answer and explanation
Correct option: C
S−1S\frac{S-1}{S}
Option C (S−1S\frac{S-1}{S}) is correct because it is derived directly from the formula for the sum of an infinite geometric series.
101 Easy
101 of 998
Application +4 | -1
If a,b,ca, b, c, dd are H.P., then
View answer and explanation
Correct option: C
ad>bca d>b c
Option C (ad>bca d>b c) is correct because it is derived from the relationship between terms in a harmonic progression, using the properties of geometric and harmonic means.
102 Easy
102 of 998
Application +4 | -1
If ∣x∣<1\left| x \right| < 1 then the sum of the series 1+2x+3x2+4x3+...∞1 + 2x + 3{x^2} + 4{x^3} + ...\infty will be
View answer and explanation
Correct option: D
1(1−x)2\frac{1}{{{{\left( {1 - x} \right)}^2}}}
Option D (1(1−x)2\frac{1}{{{{\left( {1 - x} \right)}^2}}}) is correct because the series can be expressed as the derivative of a geometric series.
103 Medium
103 of 998
Application +4 | -1
If a,b,c,da, b, c, d are positive real number such that a+b+c+d=2a+b+c+d=2, then M=(a+b)(c+d)M=(a+b)(c+d) satisfies the relation:
View answer and explanation
Correct option: A
0<M≤10 < M \leq 1
Option A ((2n+1)π2(2 n+1) \frac{\pi}{2}) is correct because the product of the matrices is a null matrix when cos⁡(θ−ϕ)=0\cos(\theta - \phi) = 0, which occurs when θ−ϕ\theta - \phi is an odd multiple of π2\frac{\pi}{2}.
104 Easy
104 of 998
Understanding +4 | -1
Consider the following statements

I. If a1,a2,…,an…{a}_1, {a}_2, \ldots, {a}_{{n}} \ldots is a sequence, then the expression a1+a2+…+an+…a_1+a_2+\ldots+a_n+\ldots is called a series.

II. Those sequences whose terms follow certain patterns are called progressions.

Choose the correct option.
View answer and explanation
Correct option: D
Both are true
Option D (Both are true) is correct because both statements accurately describe the definition of a series and a progression.
105 Easy
105 of 998
Application +4 | -1
Suppose all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6:116: 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is
View answer and explanation
Correct option: D
9
Option D (9) is correct because the common difference is found by using the given ratio of sums and the condition on the seventh term.
106 Easy
106 of 998
Application +4 | -1
If 3+log⁡5x=2log⁡25y,3 + {\log _5}x = 2{\log _{25}}y,then xx equals to
View answer and explanation
Correct option: A
y125\frac{y}{{125}}
Option A (y125\frac{y}{{125}}) is correct because by using logarithm properties, the equation can be simplified to express x in terms of y.
107 Easy
107 of 998
Application +4 | -1
If A1,A2A_1, A_2 be two A.M.'s and G1,G2G_1, G_2 be two G.M.'s between aa and bb, then A1+A2G1G2\frac{A_1+A_2}{G_1 G_2} is equal to
View answer and explanation
Correct option: C
a+bab\frac{a+b}{a b}
Option C (a+bab\frac{a+b}{ab}) is correct because, after calculating A1+A2 and G1G2, their ratio simplifies to (a+b)/ab.
108 Easy
108 of 998
Application +4 | -1
Let aa be a positive number such that the arithmetic mean of aa and 2 exceeds their geometric mean by 1. Then, the value of aa is
View answer and explanation
Correct option: D
8
Option D (8) is correct because solving the equation a+22=2a+1\frac{{a + 2}}{2} = \sqrt {2a} + 1 leads to a=8a = 8.
109 Medium
109 of 998
Application +4 | -1
{an}\left\{a_n\right\} and {bn}\left\{b_n\right\} are arithmetic progressions and their sums of the first nn terms are AnA_n and BnB_{n} respectively. If AnBn=n−12n\frac{A_n}{B_n}=\frac{n-1}{2 n} for all positive integers n,n, then the value of a3+a5+a73(b3+b9)+a4+a102(b2+b10)\frac{a_3+a_5+a_7}{3\left(b_3+b_9\right)}+\frac{a_4+a_{10}}{2\left(b_2+b_{10}\right)} is:
View answer and explanation
Correct option: D
511\frac{5}{11}
Option D (5/11) is correct because after simplifying the given ratio of sums and using the properties of arithmetic progressions, the expression evaluates to 5/11.
110 Medium
110 of 998
Analyzing +4 | -1
If 0<y<21/30 < y < {2^{1/3}} and x(y3−1)=1x\left( {{y^3} - 1} \right) = 1, then 2x+23x3+25x5+…\frac{2}{x} + \frac{2}{{3{x^3}}} + \frac{2}{{5{x^5}}} + \ldots is equal to
View answer and explanation
Correct option: A
log⁡(y32−y3)\log \left( {\frac{{{y^3}}}{{2 - {y^3}}}} \right)
Option A (log⁡(y32−y3)\log \left( {\frac{{{y^3}}}{{2 - {y^3}}}} \right)) is correct because it is derived by substituting x=1y3−1x = \frac{1}{{{y^3} - 1}} into the given series and simplifying using logarithmic properties.
111 Easy
111 of 998
Analyzing +4 | -1
Let bi>1b_i>1 for i=1,2,…,101i=1,2, \ldots, 101. Suppose log⁡eb1\log _e b_1, log⁡eb2,…,log⁡eb101\log _e b_2, \ldots, \log _e b_{101} are in Arithmetic Progression (A.P) with the common differencelog⁡e2{\log _e}2 . Suppose a1,a2,…,a101a_1, a_2, \ldots, a_{101} are in A.P. such that a1=b1a_1=b_1 and a51=b51a_{51}=b_{51}. If t=b1+b2+…+b51t=b_1+b_2+\ldots+b_{51} and s=a1+a2+…+a51s=a_1+a_2+\ldots+a_{51}, then
View answer and explanation
Correct option: B
s>ts>t and a101<b101a_{101} < b_{101}
Option B (s>ts>t and a101<b101a_{101} < b_{101}) is correct because the sum of the arithmetic progression is greater than the sum of the geometric progression, while the 101st term of the arithmetic progression is less than the 101st term of the geometric progression.
112 Easy
112 of 998
Application +4 | -1
If 1+cos⁡α+cos⁡2α+...∞=2−2,1 + \cos \alpha + {\cos ^2}\alpha + ...\infty = 2 - \sqrt 2 , then α,  (0<α<π)\alpha ,\;(0 < \alpha < \pi ) is
View answer and explanation
Correct option: D
3π4\frac{{3\pi }}{4}
Option D (3π4\frac{{3\pi }}{{4}}) is correct because it is the value of α\alpha that satisfies the given infinite geometric series equation.
113 Medium
113 of 998
Application +4 | -1
The minimum number of terms from the beginning of the series 20+2223+2513+…20 + 22\frac{2}{3} + 25\frac{1}{3} + \ldots so that the sum may exceed 1568, is
View answer and explanation
Correct option: D
29
Option D is correct because the minimum number of terms is found by solving the inequality for the sum of an arithmetic series exceeding 1568.
114 Medium
114 of 998
Application +4 | -1
If 12+22+32+…+20032=(2003)(4007)(334){1^2} + {2^2} + {3^2} + \ldots + {2003^2} = \left( {2003} \right)\left( {4007} \right)\left( {334} \right) and, 1×2003+2×2002+3×2001+…+2003×1=(2003)(334)x,1 \times 2003 + 2 \times 2002 + 3 \times 2001 + \ldots + 2003 \times 1 = \left( {2003} \right)\left( {334} \right)x, then x=x =
View answer and explanation
Correct option: A
2005
Option A (2005) is correct because by manipulating the given equations and simplifying, we can solve for x and find its value to be 2005.
115 Medium
115 of 998
Application +4 | -1
The sum of the first nn terms of the series 12+5+15+8+18+11+  ...\frac{1}{{\sqrt 2 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt {11} }} + \;... is
View answer and explanation
Correct option: A
13(3n+2−2)\frac{1}{3}\left( {\sqrt {3n + 2} - \sqrt 2 } \right)
Option A (1/3(√(3n+2) - √2)) is correct because it is the simplified sum of the series after rationalizing the denominator.
116 Easy
116 of 998
Application +4 | -1
The third term of a geometric progression is 4 . The product of the first five terms is
View answer and explanation
Correct option: B
454^5
Option B (454^5) is correct because the product of the first five terms of a GP with the third term being 4 is 454^5.
117 Easy
117 of 998
Application +4 | -1
The arithmetic mean between two numbers is AA, and SS is the sum of nn arithmetic means between these mumbers, then
View answer and explanation
Correct option: A
S=nAS=n A
Option A (S=nAS=n A) is correct because the sum of n arithmetic means between two numbers is n times the arithmetic mean of the two numbers.
118 Easy
118 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+\ldots is
View answer and explanation
Correct option: A
n2n^2
Option A (n2n^2) is correct because it represents the sum to infinity of the given series, derived using the formula for the sum of an infinite geometric progression.
119 Medium
119 of 998
Application +4 | -1
If x,y,zx,y,z are in HP, 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 it is derived by simplifying the logarithmic expression using the properties of harmonic progressions.
120 Easy
120 of 998
Application +4 | -1
Let TrT_r be the rth r^{\text {th }} term of an A.P. whose first term is a and common difference is dd. If for some positive integers m,nm, n, m≠n,Tm=1nm \neq n, T_m=\frac{1}{n} and Tn=1mT_n=\frac{1}{m}, then a−da-d equals
View answer and explanation
Correct option: D
0
Option D (0) is correct because a-d equals 0 when solving the equations derived from the given conditions of the arithmetic progression.
Revision notes and key takeaways

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

Showing 91-120 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.