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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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Showing 151-180 of 564 questions

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151 Medium
151 of 564
Understanding +4 | -1
The arithmetic mean, geometric mean and harmonic mean between two positive real quantities are themselves in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because the arithmetic mean (A), geometric mean (G), and harmonic mean (H) between two positive real quantities satisfy the relation G2 = AH, indicating they are in geometric progression.
152 Medium
152 of 564
Application +4 | -1
The sum to n\mathrm{n} terms of the series 312+512+22+712+22+32+……\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots \ldots is
View answer and explanation
Correct option: B
6nn+1\frac{6 n}{n+1}
Option B 6nn+1\frac{{6n}}{{n+1}} is correct because the series telescopes after rewriting the general term.
153 Medium
153 of 564
Application +4 | -1
The sum of 0.2+0.004+0.00006+0.0000008+…is0.2+0.004+0.00006+0.0000008+\ldots i s
View answer and explanation
Correct option: B
20009801\frac{2000}{9801}
Option B (20009801\frac{2000}{9801}) is correct because the sum of the given series, which is an arithmetico-geometric progression, is calculated correctly using the appropriate formula.
154 Medium
154 of 564
Application +4 | -1
The sum of three decreasing numbers in A.P. is 27. If −1,−1,3-1,-1,3 are added to them respectively, the resulting series is in G.P. The numbers are
View answer and explanation
Correct option: D
17,9,117,9,1
The correct answer is D (17,9,117,9,1) because these numbers satisfy both the A.P. and G.P. conditions described in the problem.
155 Medium
155 of 564
Application +4 | -1
If 19th term of a non-zero AP is zero, then its (49th term) : (29th term) is
View answer and explanation
Correct option: D
03:01
Option D (03:01) is correct because using the given condition that the 19th term is zero, the ratio of the 49th term to the 29th term is 3:1.
156 Medium
156 of 564
Analyzing +4 | -1
If ∣a∣<1\left| a \right| < 1 and ∣b∣<1,\left| b \right| < 1, then the sum of the series a(a+b)+a2(a2+b2)+a3(a3+b3)+…a\left( {a + b} \right) + {a^2}\left( {{a^2} + {b^2}} \right) + {a^3}\left( {{a^3} + {b^3}} \right) + \ldots upto ∞\infty, is
View answer and explanation
Correct option: B
a21−a2+ab1−ab\frac{{{a^2}}}{{1 - {a^2}}} + \frac{{ab}}{{1 - ab}}
Option B (a21−a2+ab1−ab\frac{{{a^2}}}{{1 - {a^2}}} + \frac{{ab}}{{1 - ab}}) is correct because the series can be split into two geometric series, and their sums can be calculated.
157 Medium
157 of 564
Application +4 | -1
If the sum of 1+1+22+1+2+33+…1+\frac{1+2}{2}+\frac{1+2+3}{3}+\ldots to nn terms is SS, then SS is equal to
View answer and explanation
Correct option: A
n(n+3)4\frac{n(n+3)}{4}
Option A (n(n+3)4\frac{n(n+3)}{4}) is correct because it represents the sum of the series 1 + (1+2)/2 + (1+2+3)/3 + ... to n terms, calculated by finding the general term and summing it.
158 Medium
158 of 564
Analyzing +4 | -1
If x,y,zx,y,z are in G.P. and x+3,y+3,z+3x + 3,y + 3,z + 3 are in H.P., then
View answer and explanation
Correct option: B
y=3y = 3
Option B (y=3y = 3) is correct because substituting y=3y=3 into the given conditions satisfies both the geometric and harmonic progression relationships.
159 Medium
159 of 564
Application +4 | -1
The sum of first 50 terms of i,−2,−3i,4,5i…i,-2,-3 i, 4,5 i \ldots is
View answer and explanation
Correct option: D
−26+25i-26+25 i
Option D (-26+25i) is correct because it is the sum of the first 50 terms of the given series.
160 Medium
160 of 564
Analyzing +4 | -1
A number consists of three digits in GP. If the sum of extreme digits is greater than twice middle digit by 1 and the sum of the left hand and middle digit is two-third of the sum of the middle and right hand digits, then the number is
View answer and explanation
Correct option: A
4,6,94,6,9
Option A (4,6,94,6,9) is the correct answer because it satisfies both given conditions for a three-digit number in GP.
161 Medium
161 of 564
Analyzing +4 | -1
If a,b,ca,b,c are three unequal positive quantities in H.P., then
View answer and explanation
Correct option: B
a5+c5>2b5{a^5} + {c^5} > 2{b^5}
Option B (a5+c5>2b5{a^5} + {c^5} > 2{b^5}) is correct. Since a,b,ca, b, c are in H.P., an+cn>2bn{a^n} + {c^n} > 2{b^n} holds for n>1n > 1, and thus a5+c5>2b5{a^5} + {c^5} > 2{b^5} is true.
162 Medium
162 of 564
Application +4 | -1
The arithmetic mean of two numbers is 183418 \frac{3}{4} and the positive square root of their product is 15 . The larger of the two numbers is
View answer and explanation
Correct option: D
30
Option D (30) is correct because solving the arithmetic mean and geometric mean equations yields 30 as the larger number.
163 Medium
163 of 564
Analyzing +4 | -1
The sum of the infinite series 1+12!+1.34!+1.3.56!+…1 + \frac{1}{{2!}} + \frac{{1.3}}{{4!}} + \frac{{1.3.5}}{{6!}} + \ldots is
View answer and explanation
Correct option: C
e\sqrt e
Option C (e\sqrt e) is correct. The given series can be rewritten as the expansion of e1/2e^{1/2}, which simplifies to e\sqrt{e}.
164 Medium
164 of 564
Application +4 | -1
If A1,A2;G1,G2A_1, A_2 ; G_1, G_2 and H1,H2H_1, H_2 be two A.M.s, G.M.s and H.M.s between two numbers respectively, then G1G2H1H2×H1+H2A1+A2=\frac{G_1 G_2}{H_1 H_2} \times \frac{H_1+H_2}{A_1+A_2}=
View answer and explanation
Correct option: A
1
The correct answer is 1 because, based on the properties of AM, GM, and HM, the expression simplifies to 1.
165 Medium
165 of 564
Application +4 | -1
If the interior angles of a polygon are in A.P. with common difference 5∘5^\circ and the smallest angle 120∘120^\circ, then the number of sides of the polygon is
View answer and explanation
Correct option: B
9\mathrm{9}
Option B (9) is correct because the number of sides of the polygon is 9, considering the constraint that the interior angles must be less than 180 degrees.
166 Medium
166 of 564
Application +4 | -1
If 10 th term of a GP is 9 and 4 th term is 4 , then its 7 th term is
View answer and explanation
Correct option: A
6
Option A (6) is correct because the 7th term of the GP can be found by using the given 10th and 4th terms.
167 Medium
167 of 564
Analyzing +4 | -1
12.2213+22.3213+23+32.4213+23+33+...+n\frac{{\frac{1}{2}.\frac{2}{2}}}{{{1^3}}} + \frac{{\frac{2}{2}.\frac{3}{2}}}{{{1^3} + {2^3}}} + \frac{{\frac{3}{2}.\frac{4}{2}}}{{{1^3} + {2^3} + {3^3}}} + ... + n terms equals
View answer and explanation
Correct option: C
(nn+1)\left( {\frac{n}{{n + 1}}} \right)
Option C (nn+1)\left( {\frac{{n}}{{n + 1}}} \right) is the correct answer because the series simplifies to a telescoping series, resulting in the given expression.
168 Medium
168 of 564
Analyzing +4 | -1
If a+beya−bey=b+ceyb−cey=c+deyc−dey\frac{a+b e^y}{a-b e^y}=\frac{b+c e^y}{b-c e^y}=\frac{c+d e^y}{c-d e^y}, then a,b,ca, b, c and dd are in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because applying componendo and dividendo leads to a constant ratio between consecutive terms, indicating a geometric progression.
169 Medium
169 of 564
Application +4 | -1
If Sn\mathrm{S}_n is the sum of the first nn terms of an A.P. and S3n=4Sn\mathrm{S}_{3 n}=4 \mathrm{S}_n, then S4nSn=\frac{\mathrm{S}_{4 n}}{\mathrm{S}_n}=
View answer and explanation
Correct option: C
6
Option B (6) is correct because the ratio of S_{4n} to S_n is 6, as determined by using the properties of arithmetic progressions.
170 Medium
170 of 564
Application +4 | -1
If the sum of four integers in AP is 24 and their product is 945 , then the numbers are
View answer and explanation
Correct option: A
3,5,7,93,5,7,9
Option A (3,5,7,93,5,7,9) is correct because the sum of these integers in AP is 24 and their product is 945.
171 Medium
171 of 564
Analyzing +4 | -1
1×22+2×32+3×42+−−−−+11(12)212×2+22×3+32×4+−−−−+112(12)=\frac{1 \times 2^2+2 \times 3^2+3 \times 4^2+----+11(12)^2}{1^2 \times 2+2^2 \times 3+3^2 \times 4+----+11^2(12)}=?
View answer and explanation
Correct option: A
1917\frac{19}{17}
Option D (3x2−20x+16=03 x^2-20 x+16=0) is correct because after simplifying the given equation and finding the new roots, the resulting quadratic equation is 3x2−20x+16=03x^2 - 20x + 16 = 0.
172 Medium
172 of 564
Application +4 | -1
Sum to n terms of the series 11  .    2  .    3  .    4+12  .    3  .    4  .    5  +13  .    4  .    5  .  6  +…,\frac{1}{{1\;.\;\;2\;.\;\;3\;.\;\;4}} + \frac{1}{{2\;.\;\;3\;.\;\;4\;.\;\;5\;}} + \frac{1}{{3\;.\;\;4\;.\;\;5\;.\;6\;}} + \ldots , is
View answer and explanation
Correct option: D
n3+6n2+11n18(n+1)(n+2)(n+3)\frac{{{n^3} + 6{n^2} + 11n}}{{18\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}
Option D (n3+6n2+11n18(n+1)(n+2)(n+3)\frac{{{n^3} + 6{n^2} + 11n}}{{18\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}) is correct because it represents the sum to n terms of the given series after using partial fractions.
173 Medium
173 of 564
Analyzing +4 | -1
If k≠0k \ne 0 is a constant and n∈Nn \in N, then, lim⁡n→∞  {n!(k  n)n}\lim\limits_{n \to \infty } \,\,\left\{ {\frac{{n!}}{{{{\left( {k\;n} \right)}^n}}}} \right\} is equal to
View answer and explanation
Correct option: D
1ke\frac{1}{{ke}}
Option D (1ke\frac{1}{{ke}}) is correct because the limit is evaluated using Stirling's approximation and properties of logarithms and integrals.
174 Medium
174 of 564
Application +4 | -1
Let G\mathrm{G} be the geometric mean of two positive numbers a and b\mathrm{b}, and M\mathrm{M} be the arithmetic mean of 1a\frac{1}{\mathrm{a}} and 1b\frac{1}{\mathrm{b}}. If 1M:G\frac{1}{\mathrm{M}}: \mathrm{G} is 4:54: 5, then a:ba: b can be:
View answer and explanation
Correct option: A
1:41: 4
Option A (1:4) is correct because it satisfies the given condition 1M:G=4:5\frac{1}{M}: G = 4:5 after simplifying the geometric and arithmetic means.
175 Medium
175 of 564
Analyzing +4 | -1
The sum of

12⋅2213+22.3213+23+32.4213+23+33+...\frac{{\frac{1}{2} \cdot \frac{2}{2}}}{{{1^3}}} + \frac{{\frac{2}{2}.\frac{3}{2}}}{{{1^3} + {2^3}}} + \frac{{\frac{3}{2}.\frac{4}{2}}}{{{1^3} + {2^3} + {3^3}}} + ... upto nn terms is equal to
View answer and explanation
Correct option: B
nn+1\frac{n}{{n + 1}}
Option B (nn+1\frac{n}{{n + 1}}) is correct because the series telescopes to this result after simplification and partial fraction decomposition.
176 Medium
176 of 564
Application +4 | -1
If x,yx, y and zz are in HP, where z>y>xz>y>x. Then, value of log⁡(x+z)+log⁡(x−2y+z)\log (x+z)+\log (x-2 y+z) is
View answer and explanation
Correct option: B
2log⁡(z−x)2 \log (z-x)
Option B (2log⁡(z−x)2 \log (z-x)) is correct because using the properties of Harmonic Progression, we can simplify the expression to obtain the result.
177 Medium
177 of 564
Analyzing +4 | -1
The sum of (x+2)n−1+(x+2)n−2(x+1)+(x+2)n−3(x+2)^{n-1}+(x+2)^{n-2}(x+1)+(x+2)^{n-3} (x+1)2+…+(x+1)n−1(x+1)^2+\ldots+(x+1)^{n-1} is equal to
View answer and explanation
Correct option: C
(x+2)x−(x+1)n(x+2)^x-(x+1)^n
Option C ((x+2)x−(x+1)n(x+2)^x-(x+1)^n) is the correct answer because the given series is a geometric progression sum that simplifies to the expression.
178 Medium
178 of 564
Application +4 | -1
If the AM is twice the GM of the numbers aa and bb, then a:ba: b will be
View answer and explanation
Correct option: B
2+32−3\frac{2+\sqrt{3}}{2-\sqrt{3}}
Option B (2+32−3\frac{2+\sqrt{3}}{2-\sqrt{3}}) is correct because solving the given condition AM = 2GM leads to the ratio a:b=(2+3):(2−3)a:b = (2+\sqrt{3}):(2-\sqrt{3}).
179 Medium
179 of 564
Analyzing +4 | -1
If a,b,ca, b, c are in AP and ∣a∣,∣b∣,∣c∣<1|a|,|b|,|c|<1 and x=1+a+a2+…+∞x=1+a+a^2+\ldots+\infty, y=1+b+b2+…+∞y=1+b+b^2+\ldots+\infty, z=1+c+c2+…+∞z=1+c+c^2+\ldots+\infty
Then, x,y,zx, y, z will be in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because if a, b, c are in AP, then 1-a, 1-b, 1-c are also in AP. Since x, y, z are the reciprocals of these terms, they are in HP.
180 Medium
180 of 564
Application +4 | -1
Let SnS_n denote the sum of first nn terms of an arithmetic progression. If S20=790S_{20}=790 and S10=145S_{10}=145, then S14−S5S_{14}-S_5 is :
View answer and explanation
Correct option: D
395
The correct answer is D (395) because it is the result of calculating the difference between the sum of the first 14 terms and the sum of the first 5 terms of the arithmetic progression.
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