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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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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.

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271 Medium
271 of 564
Application +4 | -1
Ankur appeared in an examination which has 5 subjects, out of five, in four subjects he got 90,70,75,6590,70,75,65 marks respectively. The minimum & maximum marks he should score in fifth subject so that the average mark is greater than or equal to 70 and less than or equal to 75 is
View answer and explanation
Correct option: C
50,75
The correct answer is A (55,75) because the minimum and maximum marks are calculated by setting up inequalities based on the given average range.
272 Medium
272 of 564
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.
273 Medium
273 of 564
Analyzing +4 | -1
If (10)9+2(11)1(10)8+3(11)2(10)7+…+(10)(11)9(10)^9+2(11)^1(10)^8+3(11)^2(10)^7+\ldots+(10)(11)^9 =k(10)9=k(10)^9, then kk is equal to
View answer and explanation
Correct option: A
100
Option A (100) is correct because the given series can be simplified and summed to find the value of k, which is 100.
274 Medium
274 of 564
Analyzing +4 | -1
Let a,b,c∈Ra, b, c \in R. If f(x)=x2+bx+cf(x)=x^{2}+b x+c is such that a+b+c=3a+b+c=3 and f(x+y)=f(x)+f(y)+xy,∀x,y∈Rf(x+y)=f(x)+f(y)+x y, \forall x, y \in R, then ∑n=110f(n)\sum_{n=1}^{10} f(n) is equal to:
View answer and explanation
Correct option: B
330
Option B (330) is correct because by using the given functional equation and the initial condition, we can determine the function and then calculate the sum.
275 Medium
275 of 564
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.
276 Medium
276 of 564
Application +4 | -1
If an+1+bn+1an+bn\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} is the harmonic mean between aa and bb, then the value of nn is
View answer and explanation
Correct option: B
-1
Option B (-1) is correct because substituting the harmonic mean formula and simplifying leads to n = -1.
277 Medium
277 of 564
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).
278 Medium
278 of 564
Analyzing +4 | -1
If a,b,ca, b, c are three distinct real numbers are in G.P and a+b+c=xba+b+c=x b then Which of the following statements are true

Statement-I : xx can not take the values between −1-1 and 2

Statement-II : The least positive integral value of xx is 3
View answer and explanation
Correct option: C
I is true, II is false
Option C (I is true, II is false) is correct because the value of x can not take values between -1 and 3, and the least positive integral value of x is 3.
279 Medium
279 of 564
Application +4 | -1
The nnth term of a GP is 128 and the sum to its nn terms is 255 . If its common ratio is 2 , then its first term is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by using the formula for the sum of a GP and the nth term, we can solve for the first term, a, which equals 1.
280 Medium
280 of 564
Application +4 | -1
If S=∑n=2∞  nC23n−2n  !,S = \sum \limits_{n = 2}^\infty {\;^n}{C_2}\frac{{{3^{n - 2}}}}{{n\;!}}, then SS equals
View answer and explanation
Correct option: B
12e3\frac{1}{2}{e^3}
Option B (12e3\frac{1}{2}{e^3}) is correct because the series can be simplified and recognized as a Taylor series expansion of the exponential function.
281 Medium
281 of 564
Application +4 | -1
The sum of nn terms of the following series 1.2+2.3+3.4+4.5+…1.2+2.3+3.4+4.5+\ldots shall be
View answer and explanation
Correct option: B
13n(n+1)(n+2)\frac{1}{3} n(n+1)(n+2)
Option B (13n(n+1)(n+2)\frac{1}{3}n(n+1)(n+2)) is correct because it represents the sum of the series 1.2 + 2.3 + 3.4 + ... to n terms, obtained by summing the general term n(n+1).
282 Medium
282 of 564
Analyzing +4 | -1
Let Sn=113+1+213+23+1+2+313+23+33+…S_{n}=\frac{1}{1^{3}}+\frac{1+2}{1^{3}+2^{3}}+\frac{1+2+3}{1^{3}+2^{3}+3^{3}}+\ldots. +1+2+……+n13+23+……+n3+\frac{1+2+\ldots \ldots+n}{1^{3}+2^{3}+\ldots \ldots+n^{3}}, If 100Sn=n100 \mathrm{S}_{n}=n, then n\mathrm{n} is equal to:
View answer and explanation
Correct option: A
199
Option A (199) is correct because simplifying the series allows us to find a closed form and solve for n.
283 Medium
283 of 564
Application +4 | -1
If Sn=16.11+111.16+116.21+...to  n{S_n} = \frac{1}{{6.11}} + \frac{1}{{11.16}} + \frac{1}{{16.21}} + ...\mathrm{to}\;n terms, then 6Sn6{S_n} equals
View answer and explanation
Correct option: B
n(5n+6)\frac{n}{{\left( {5n + 6} \right)}}
Option B is correct because the series can be expressed as a telescoping series, allowing for simplification and direct calculation of the sum.
284 Medium
284 of 564
Application +4 | -1
If a,b,ca, b, c and dd are in HP, then ab+bc+cda b+b c+c d is equal to
View answer and explanation
Correct option: C
3ad3 a d
Option C (3ad3ad) is correct because, in a Harmonic Progression, the reciprocals are in AP, and the given expression can be derived using the AP properties.
285 Medium
285 of 564
Application +4 | -1
If it is given that 114+124+134+…to∞=π490,\frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + \ldots \mathrm{to }\infty = \frac{{{\mathrm{\pi }^4}}}{{90}},, then the value of 114+134+154+…to∞\frac{1}{{{1^4}}} + \frac{1}{{{3^4}}} + \frac{1}{{{5^4}}} + \ldots \mathrm{to }\infty is equal to
View answer and explanation
Correct option: A
π496\frac{{{\pi ^4}}}{{96}}
Option A (π496\frac{{{\pi ^4}}}{{96}}) is correct because it is derived by subtracting the sum of the series with even terms from the total sum.
286 Medium
286 of 564
Application +4 | -1
If a,ba, b and cc are in APA P, then a+1bc,b+1caa+\frac{1}{b c}, b+\frac{1}{c a} and c+1abc+\frac{1}{a b} are in
View answer and explanation
Correct option: B
APA P
Option B (APAP) is correct because if a,b,ca, b, c are in AP, then a+1bc,b+1caa+\frac{1}{bc}, b+\frac{1}{ca} and c+1abc+\frac{1}{ab} are also in AP.
287 Medium
287 of 564
Application +4 | -1
If xx is added to each of numbers 3,9,213,9,21 so that the resulting numbers may be in GP, then the value of xx will be
View answer and explanation
Correct option: A
3
Option A (3) is correct because adding 3 to each number results in a geometric progression.
288 Medium
288 of 564
Application +4 | -1
If S1,S2,S3{S_1},{S_2},{S_3} be the sum of n,2n,3nn,2n,3n nn terms respectively of an A.P., then
View answer and explanation
Correct option: C
S3=3(S2−S1){S_3} = 3\left( {{S_2} - {S_1}} \right)
Option C (S3=3(S2−S1){S_3} = 3\left( {{S_2} - {S_1}} \right)) is correct because it represents the relationship between the sums of n, 2n, and 3n terms of an arithmetic progression.
289 Medium
289 of 564
Application +4 | -1
The value of log⁡22222  ,  is{\log _{\sqrt 2 }}\sqrt {2\sqrt {2\sqrt {2\sqrt 2 } } } \;,\;\mathrm{is}
View answer and explanation
Correct option: C
158\frac{{15}}{8}
Option C (158\frac{{15}}{8}) is correct because it is obtained by simplifying the nested radicals and applying logarithm properties.
290 Medium
290 of 564
Analyzing +4 | -1
The sum upto (2n+1)\left( {2n + 1} \right) terms of the series a2−(a+d)2+(a+2d)2−(a+3d)2+…,{a^2} - {\left( {a + d} \right)^2} + {\left( {a + 2d} \right)^2} - {\left( {a + 3d} \right)^2} + \ldots , is
View answer and explanation
Correct option: D
a2+2nad+n(2n+1)d2{a^2} + 2nad + n\left( {2n + 1} \right){d^2}
Option C (a2+2nad+n(2n+1)d2{a^2} + 2nad + n\left( {2n + 1} \right){d^2}) is correct because the sum of the series can be simplified and expressed in the given form.
291 Medium
291 of 564
Application +4 | -1
If a,b,ca,b,c are in A.P., then a+1bc,b+1ca,c+1aba + \frac{1}{{bc}},b + \frac{1}{{ca}},c + \frac{1}{{ab}} are in
View answer and explanation
Correct option: A
A.P.
Option A (A.P.) is correct because if a, b, c are in A.P., then a + 1/bc, b + 1/ca, c + 1/ab are also in A.P.
292 Medium
292 of 564
Application +4 | -1
If 1a,1b\frac{1}{a}, \frac{1}{b} and 1c\frac{1}{c} are in AP, then (1a+1b−1c)\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right) (1b+1c−1a)\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right) is equal to
View answer and explanation
Correct option: A
4ac−3b2\frac{4}{a c}-\frac{3}{b^2}
Option A (4ac−3b2\frac{4}{ac} - \frac{3}{b^2}) is correct because it is obtained by substituting the AP condition and simplifying the given expression.
293 Medium
293 of 564
Application +4 | -1
If a1,a2…,ana_1, a_2 \ldots, a_n are positive real numbers whose product is a fixed number c\mathrm{c}, then the minimum value of a1+a2+……+an−1+2ana_1+a_2+\ldots \ldots+a_{n-1}+2 a_n is
View answer and explanation
Correct option: A
n(2c)1/nn(2 c)^{1 / n}
Option A (n(2c)1/nn(2 c)^{1 / n}) is correct because it applies the AM-GM inequality to find the minimum value of the given expression.
294 Medium
294 of 564
Analyzing +4 | -1
If log⁡(x+z)+log⁡(x+z−2y)=2log⁡(x−z)\log (x+z)+\log (x+z-2 y)=2 \log (x-z), then x,yx, y and zz are in
View answer and explanation
Correct option: C
HP\mathrm{HP}
Option HP is correct because by simplifying the given logarithmic equation, we can show that x, y, and z are in HP.
295 Medium
295 of 564
Application +4 | -1
If x=(2+1)3−(2−1)3x=\sqrt[3]{(\sqrt{2}+1)}-\sqrt[3]{(\sqrt{2}-1)}; then x3+3x=x^{3}+3 x=
View answer and explanation
Correct option: A
2
Option A is correct because the expression simplifies to 2 by cubing x and using the identity a3−b3=(a−b)3+3ab(a−b)a^3 - b^3 = (a-b)^3 + 3ab(a-b).
296 Medium
296 of 564
Application +4 | -1
If cos⁡(x−y),cos⁡x\cos (x-y), \cos x and cos⁡(x+y)\cos (x+y) are in HP, then cos⁡xsec⁡(y2)\cos x \sec \left(\frac{y}{2}\right) is equal to
View answer and explanation
Correct option: A
±2\pm \sqrt{2}
Option A (±2\pm \sqrt{{2}}) is correct because by using the Harmonic Progression (HP) condition and trigonometric identities, we can derive that cos⁡xsec⁡(y2)=±2\cos x \sec \left(\frac{{y}}{{2}}\right) = \pm \sqrt{{2}}.
297 Medium
297 of 564
Application +4 | -1
sin⁡−1(a−a23+a39+….)+cos⁡−1(1+b+b2+…)=π2\sin ^{-1}\left(a-\frac{a^2}{3}+\frac{a^3}{9}+\ldots.\right)+\cos ^{-1}\left(1+b+b^2+\ldots\right)=\frac{\pi}{2} when
View answer and explanation
Correct option: B
a=1  &  b=−13a=1 \,\, \& \,\, b=-\frac{1}{3}
Option B (a=1&b=−13a=1 \& b=-\frac{1}{3}) is correct because it satisfies the condition that the two series are equal and converge within their respective domains.
298 Medium
298 of 564
Application +4 | -1
If 32x+1.4x−1=36,{3^{2x + 1}}{.4^{x - 1}} = 36, then x=x =
View answer and explanation
Correct option: A
log⁡3648{\log _{36}}48
Option B (log⁡4836{\log _{48}}36) is correct because the equation can be simplified and solved for x using logarithms.
299 Medium
299 of 564
Application +4 | -1
The sum of the series 1+3x+6x2+10x3+...∞1 + 3x + 6{x^2} + 10{x^3} + ...\infty will be
View answer and explanation
Correct option: D
1(1−x)3\frac{1}{{{{\left( {1 - x} \right)}^3}}}
Option D (1(1−x)3\frac{1}{{{{\left( {1 - x} \right)}^3}}}) is correct because it is derived by using the method of difference to simplify the series.
300 Medium
300 of 564
Application +4 | -1
If log⁡10(12x+x−1)=x(log⁡105−1)\log _{10}\left(\frac{1}{2^x+x-1}\right)=x\left(\log _{10} 5-1\right) then xx is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because substituting x=1 into the equation satisfies the given logarithmic equation.
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