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 631-660 of 998 questions

Preparing practice and saved progress…

631 Easy
631 of 998
Analyzing +4 | -1
Let x,yx, y be positive real numbers and m,nm, n positive integers. The maximum value of the expression xmyn(1+x2m)(1+y2n)\frac{{{x^m}{y^n}}}{{(1 + {x^{2m}})(1 + {y^{2n}})}} is
View answer and explanation
Correct option: C
14\frac{1}{4}
Option C (14\frac{1}{4}) is correct because the maximum value of the expression is achieved when xm=1x^m = 1 and yn=1y^n = 1, resulting in 14\frac{1}{4} after applying AM-GM inequality.
632 Easy
632 of 998
Analyzing +4 | -1
The sum of the series 1+43+109+2827+…1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+\ldots upto nn terms is:
View answer and explanation
Correct option: D
n+12−12⋅3n−1n+\frac{1}{2}-\frac{1}{2 \cdot 3^{n-1}}
Option D (n+12−12⋅3n−1n+\frac{1}{2}-\frac{1}{2 \cdot 3^{n-1}}) is correct because it represents the sum of the given series up to n terms, derived by splitting the series and using the formula for the sum of a geometric progression.
633 Easy
633 of 998
Application +4 | -1
The sum of the series 13×7+17×11+111×15+...\frac{1}{{3 \times 7}} + \frac{1}{{7 \times 11}} + \frac{1}{{11 \times 15}} + ... is
View answer and explanation
Correct option: D
112\frac{1}{{12}}
Option D (112\frac{1}{{12}}) is correct because it is the sum of the infinite series, calculated by expressing each term as a difference and observing the telescoping nature of the series.
634 Easy
634 of 998
Application +4 | -1
Suppose a1=45,a2=41a_1=45, a_2=41 and ak=2ak−1−ak−2 ∀ k≥3a_k=2 a_{k-1}-a_{k-2} \, \forall \, k \geq 3, then, value of

S=112[a12−a22+a32−a42+…+a112−a122]S=\frac{1}{12}\left[a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{11}^2-a_{12}^2\right] is
View answer and explanation
Correct option: D
92
Option D (92) is correct because the given recurrence relation implies an arithmetic progression, and the sum is calculated correctly using the properties of an A.P.
635 Easy
635 of 998
Analyzing +4 | -1
If a+bxa−bx=b+cxb−cx=c+dxc−dx(x≠0),    then  a,  b,  c,  d\frac{{a + bx}}{{a - bx}} = \frac{{b + cx}}{{b - cx}} = \frac{{c + dx}}{{c - dx}}\left( {x \ne 0} \right),\;\;\mathrm{then}\;a,\;b,\;c,\;d are in
View answer and explanation
Correct option: B
GP
Option B (GP) is the correct answer because applying componendo and dividendo leads to the conclusion that a, b, c, and d form a geometric progression.
636 Easy
636 of 998
Understanding +4 | -1
If SS is the sum of an infinite GP, the first term aa, then the common ratio rr is given by
View answer and explanation
Correct option: B
S−aS\frac{{S - a}}{S}
Option B (S−aS\frac{{S - a}}{S}) is correct because the common ratio can be derived from the formula for the sum of an infinite geometric progression.
637 Easy
637 of 998
Analyzing +4 | -1
The sum 1+2.3+3.32+....10.391 + 2.3 + {3.3^2} + {....10.3^9} is equal to
View answer and explanation
Correct option: B
19.310+14\frac{{{{19.3}^{10}} + 1}}{4}
Option B (19.310+14\frac{{{{19.3}^{10}} + 1}}{4}) is correct because by using the method of differences to sum the series, the result can be obtained.
638 Easy
638 of 998
Application +4 | -1
If the sum of second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of its first 50 terms is
View answer and explanation
Correct option: D
126(350−1)\frac{1}{26}\left(3^{50}-1\right)
Option B (2+32+\sqrt{3}) is correct because it is the common ratio that satisfies the given conditions of the increasing G.P. and the resulting A.P.
639 Easy
639 of 998
Application +4 | -1
The sum of all natural numbers less than 200 that are divisible neither by 3 nor by 5 , is
View answer and explanation
Correct option: B
10732
Option B (10732) is correct because it represents the sum of natural numbers less than 200 that are not divisible by 3 or 5, calculated using inclusion-exclusion principle.
640 Easy
640 of 998
Application +4 | -1
If a,b,ca,b,c are in G.P., then log⁡aλ,log⁡bλ,log⁡cλ{\log _a}\lambda ,{\log _b}\lambda ,{\log _c}\lambda are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the logarithms form a Harmonic Progression since a, b, c are in GP.
641 Easy
641 of 998
Application +4 | -1
If the ratio of sums to nn terms of two A.P.'s is (5n+7):(3n+2)(5 n+7):(3 n+2), then the ratio of their 17th terms is
View answer and explanation
Correct option: B
172:101172: 101
Option A (172: 101) is correct because it is the ratio of the 17th terms of the two arithmetic progressions, calculated using the given ratio of their sums.
642 Easy
642 of 998
Application +4 | -1
If nn geometric means be inserted between aa and bb, then the nnth geometric mean will be
View answer and explanation
Correct option: C
a(ba)nn+1a{\left( {\frac{b}{a}} \right)^{\frac{n}{{n + 1}}}}
Option C (a(ba)nn+1a{\left( {\frac{b}{a}} \right)^{\frac{n}{{n + 1}}}}) is correct because it represents the nth geometric mean when n geometric means are inserted between a and b.
643 Easy
643 of 998
Application +4 | -1
If x=−2,x = - 2, then the value of log⁡4(x24)−2log⁡44(x4),{\log _4}\left( {\frac{{{x^2}}}{4}} \right) - 2{\log _4}4\left( {{x^4}} \right), is
View answer and explanation
Correct option: C
−6- 6
Option C is correct because by substituting x = -2 into the logarithmic expression and simplifying using logarithmic properties, the value is found to be -6.
644 Easy
644 of 998
Analyzing +4 | -1
If x>1,y>1,z>1x>1, y>1, z>1 are in G.P., then 11+ln⁡x,11+ln⁡y,11+ln⁡z\frac{1}{1+\ln x}, \frac{1}{1+\ln y}, \frac{1}{1+\ln z} are in
View answer and explanation
Correct option: B
HP
Option B (HP) is correct because if x, y, z are in GP, then ln(x), ln(y), ln(z) are in AP, and 1/(1+ln(x)), 1/(1+ln(y)), 1/(1+ln(z)) are in HP.
645 Easy
645 of 998
Application +4 | -1
The sum to infinity of the GPG P −54,516,−564,…\frac{-5}{4}, \frac{5}{16}, \frac{-5}{64}, \ldots, is
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because the sum to infinity of the given GP is calculated using the formula S = a/(1-r) where a = -5/4 and r = -1/4.

Explore Other Chapters in MATHEMATICS

View All →
DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
646 Easy
646 of 998
Application +4 | -1
The first and last term of an A.P. are aa and ll respectively. If SS be the sum of all the terms of the A.P., then the common difference is
View answer and explanation
Correct option: A
l2−a22S−(l+a)\frac{l^2-a^2}{2 S-(l+a)}
Option A is correct because the common difference is derived by using the formulas for the last term and the sum of an A.P.
647 Easy
647 of 998
Application +4 | -1
If 9 AM’s and HM’s are inserted between the 2 and 3 and if the harmonic mean HH is corresponding to arithmetic mean A,  then  A+6HA,\;\mathrm{then}\;A + \frac{6}{H} is equal to
View answer and explanation
Correct option: C
5
Option C (5) is correct because by inserting 9 arithmetic and harmonic means between 2 and 3, and using the properties of AP and HP, we find that A+6H=5A + \frac{6}{H} = 5.
648 Easy
648 of 998
Application +4 | -1
If log⁡2(a+b)+log⁡2(c+d)≥4\log _2(a+b)+\log _2(c+d) \geq 4. Then the minimum value of the expression a+b+c+da+b+c+d is
View answer and explanation
Correct option: C
8
Option C, 8, is the correct minimum value because applying AM-GM inequality to (a+b)(a+b) and (c+d)(c+d) with the given logarithmic condition yields a+b+c+d≥8a+b+c+d \geq 8.
649 Easy
649 of 998
Analyzing +4 | -1
The value of 2−12+3−224+52−762+17−12216+…+…+ad.  inf.,\frac{{\sqrt 2 - 1}}{{\sqrt 2 }} + \frac{{3 - 2\sqrt 2 }}{4} + \frac{{5\sqrt 2 - 7}}{{6\sqrt 2 }} + \frac{{17 - 12\sqrt 2 }}{{16}} + \ldots + \ldots + \mathrm{ad}.\mathrm{\;inf}., is
View answer and explanation
Correct option: B
log⁡e2{\log _e}\sqrt 2
Option B is correct because the given series can be rewritten as a logarithmic series, and its sum converges to log⁡e2{\log _e}\sqrt 2.
650 Easy
650 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+….. , is
View answer and explanation
Correct option: A
n2n^2
Option A (n2n^2) is correct because it is the sum to infinity of the given series, derived by using the arithmetico-geometric progression formula.
651 Easy
651 of 998
Application +4 | -1
If a1/x=b1/y=c1/za^{1 / x}=b^{1 / y}=c^{1 / z} and b2=acb^{2}=a c then x+z=x+z=
View answer and explanation
Correct option: B
2y2 \mathrm{y}
Option B (2y) is correct because substituting a=kxa=k^x, b=kyb=k^y, and c=kzc=k^z into b2=acb^2=ac leads to x+z=2yx+z=2y.
652 Easy
652 of 998
Application +4 | -1
Let a1,a2,a3,……,an,…a_1, a_2, a_3, \ldots \ldots, a_n, \ldots be in A.P. If a3+a7+a11+a15=72a_3+a_7+a_{11}+a_{15}=72. Then the sum of its first 17 terms is equal to
View answer and explanation
Correct option: B
306
Option B (306) is correct because the sum of the first 17 terms is calculated using the given condition and the properties of an A.P.
653 Easy
653 of 998
Application +4 | -1
If the 2nd, 5th and 9th terms of a non-constant AP are in GP, then the common ratio of this GP is
View answer and explanation
Correct option: B
43\frac{4}{3}
The correct answer is B (43\frac{4}{3}) because the common ratio is calculated using the given condition that the 2nd, 5th, and 9th terms are in GP.
654 Easy
654 of 998
Application +4 | -1
The 5th term of the sequence 109,13203,23,…  \frac{{10}}{9},\frac{1}{3}\frac{{\sqrt {20} }}{3},\frac{2}{3}, \ldots \; is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option D (25\frac{2}{5}) is correct because it is the 5th term of the given geometric progression, calculated using the formula for the nth term of a GP.
655 Easy
655 of 998
Application +4 | -1
The number of terms of the G.P. 3,  32,  34, ...3,\,\,\frac{3}{2},\,\,\frac{3}{4},\,... needed to obtain a sum of 3069512\frac{3069}{512} is:
View answer and explanation
Correct option: B
10
Option B (10) is correct because by using the geometric progression sum formula, we can find the number of terms needed to obtain a sum of 3069512\frac{3069}{512}.
656 Easy
656 of 998
Application +4 | -1
The interior angles of a polygon are in AP. The smallest angle is 120∘120^{\circ} and the common difference is 5∘5^{\circ}. The number of sides of the polygon is
View answer and explanation
Correct option: B
9
Option B (9) is correct because solving the equation for the sum of interior angles of a polygon yields n = 9.
657 Easy
657 of 998
Application +4 | -1
If log⁡xa−b=log⁡yb−c=log⁡zc−a,\frac{{\log x}}{{a - b}} = \frac{{\log y}}{{b - c}} = \frac{{\log z}}{{c - a}}, then xyzxyz is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because when the given logarithmic ratios are equal, it implies that xyz=1xyz = 1.
658 Easy
658 of 998
Application +4 | -1
If x=∑n=0∞(−1)ntan⁡2nθx=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta and y=∑n=0∞cos⁡2nθy=\sum_{n=0}^{\infty} \cos ^{2 n} \theta, for 0<θ<π40<\theta<\frac{\pi}{4}, then:
View answer and explanation
Correct option: B
y(1−x)=1y(1-x)=1
Option B (y(1-x)=1) is correct because it is derived by expressing x and y as infinite geometric series and then finding a relationship between them.
659 Easy
659 of 998
Analyzing +4 | -1
If ∑r=1nr,103∑r=1nr2,∑r=1nr3\sum\limits_{{r=1}}^n r, \frac{\sqrt{10}}{3} \sum\limits_{{r=1}}^n r^2, \sum\limits_{{r=1}}^n r^3 are in G.P., then the whice of nn, is
View answer and explanation
Correct option: C
4
Option C (4) is correct because it satisfies the given equation derived from the condition that the sum of r, sum of r squared, and sum of r cubed are in G.P.
660 Easy
660 of 998
Analyzing +4 | -1
The value of ax−bya^x-b^y is (where x=log⁡abx=\sqrt{\log _a b} and y=log⁡ba,a>0,b>0y=\sqrt{\log _b a}, a>0, b>0 and a,b≠1)a, b \neq 1)
View answer and explanation
Correct option: C
0
Option C (0) is correct because by substituting the values of x and y, we find that a^x = b^y, thus a^x - b^y = 0.
Revision notes and key takeaways

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

Showing 631-660 of 998 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - SEQUENCE OF SERIES MCQs (All) 998 MCQs → ENGINEERING MATHEMATICS - SEQUENCE OF SERIES MCQs (Hard) 37 MCQs → ENGINEERING MATHEMATICS - SEQUENCE OF SERIES MCQs (Medium) 564 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs → PROBABILITY 1146 MCQs

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:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

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.

Continue with Free Account →