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 661-690 of 998 questions

Preparing practice and saved progress…

661 Easy
661 of 998
Analyzing +4 | -1
Let b0,b1,b2,…b_0, b_1, b_2, \ldots be a sequence, where b0=12b_0=\frac{1}{2} and bn=bn−1−bn−2bn−1+bn−2∀n≥1b_n=\frac{b_{n-1}-b_{n-2}}{b_{n-1}+b_{n-2}} \forall n \geq 1.then the value of bn∏r=2n+1(1−br1+br)b_n \prod_{r=2}^{n+1}\left(\frac{1-b_r}{1+b_r}\right) equals
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the product telescopes, leaving only the initial term b0=12b_0 = \frac{1}{2}.
662 Easy
662 of 998
Application +4 | -1
The value of 5log⁡577log⁡75{5^{\sqrt {{{\log }_5}7} }}{7^{\sqrt {{{\log }_7}5} }} is
View answer and explanation
Correct option: C
0
Option D (None of these) is correct because the expression simplifies to zero, which is not among the given options. The correct answer should be 0.
663 Easy
663 of 998
Understanding +4 | -1
Non zero equal numbers are in
View answer and explanation
Correct option: C
A.P., G.P. and H.P.
Option C (A.P., G.P. and H.P.) is correct because non-zero equal numbers satisfy the conditions for arithmetic, geometric, and harmonic progressions.
664 Easy
664 of 998
Application +4 | -1
Three positive numbers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is
View answer and explanation
Correct option: D
2+32 + \sqrt 3
Option D (2+32 + \sqrt 3) is correct because solving the quadratic equation derived from the AP condition yields this value for the common ratio, considering the GP is increasing.
665 Easy
665 of 998
Understanding +4 | -1
The value of ∑i=1n∑j=1i∑k=1j1\sum\limits_{i=1}^n \sum\limits_{j=1}^i \sum\limits_{k=1}^j 1, is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the correct result is n(n+1)(n+2)6\frac{{n(n+1)(n+2)}}{{6}}, which doesn't match any of the given summations.
666 Easy
666 of 998
Analyzing +4 | -1
The value of 21/4.41/8.81/16…∞{2^{1/4}}{.4^{1/8}}{.8^{1/16}} \ldots \infty is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the infinite product simplifies to 2 after evaluating the exponent.
667 Easy
667 of 998
Understanding +4 | -1
The sum of the series 11.2+12.3+13.4+…….+1n(n+1)\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots \ldots .+\frac{1}{n(n+1)}, is
View answer and explanation
Correct option: B
1−1n+11-\frac{1}{n+1}
Option B (1−1n+11-\frac{{1}}{{n+1}}) is correct because the series telescopes, leaving only the first and last terms after cancellation.
668 Easy
668 of 998
Analyzing +4 | -1
A set of consecutive positive integers starting with one is written on blackboard.
One number erased and then the A.M of the remaining numbers is found to be 3571735 \frac{7}{17}. The erased number is
View answer and explanation
Correct option: B
7
Option B (7) is correct because it satisfies the condition that the erased number must be an integer between 1 and n, and the average of the remaining numbers is 35 7/17.
669 Easy
669 of 998
Application +4 | -1
Sum of the series ∑r=1nr(r+1)!\sum\limits_{{r=1}}^n \frac{r}{(r+1) !} is
View answer and explanation
Correct option: B
1−1(n+1)!1-\frac{1}{(n+1) !}
Option B (1−1(n+1)!1-\frac{1}{(n+1) !}) is correct because the sum of the series can be simplified using partial fractions.
670 Easy
670 of 998
Analyzing +4 | -1
If log⁡2,log⁡(2x−1)\log 2,\log \left( {{2^x} - 1} \right) and log⁡(2x+3)\log \left( {{2^x} + 3} \right) are in A.P., then 2,  2x−1,  2x+32,\;{2^x} - 1,\;{2^x} + 3 are in
View answer and explanation
Correct option: C
G.P.
Option C (G.P.) is correct because if the logarithms are in A.P., then the original terms are in G.P.
671 Easy
671 of 998
Application +4 | -1
If (1−y)(1+2x+4x2+8x3+16x4+32x5)=1−y6,(y≠1)(1-y)\left(1+2 x+4 x^2+8 x^3+16 x^4+32 x^5\right)=1-y^6,(y \neq 1), then a value y/xy / x is
View answer and explanation
Correct option: B
2
Option A (2) is correct because, from the given equation, a possible value for y/x is 2.
672 Easy
672 of 998
Application +4 | -1
If xx satisfies the inequalities x+7<2x+3x+7<2 x+3 and 2x+4<5x+32 x+4<5 x+3, then xx lies in the interval
View answer and explanation
Correct option: C
(4,∞)(4, \infty)
Option C ((4,∞)(4, \infty)) is correct because it represents the intersection of the solution sets for both inequalities: x+7<2x+3x+7<2x+3 and 2x+4<5x+32x+4<5x+3.
673 Easy
673 of 998
Application +4 | -1
The value of log⁡ba×log⁡cb×log⁡ac,{\log _b}a \times {\log _c}b \times {\log _a}c, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the expression log⁡ba×log⁡cb×log⁡ac{\log _b}a \times {\log _c}b \times {\log _a}c simplifies to 1 using the change of base formula for logarithms.
674 Easy
674 of 998
Analyzing +4 | -1
Consider the sequence a1,a2,a3,…a_1, a_2, a_3, \ldots such that a1=1,a2=2a_1=1, a_2=2 and an+2=2an+1+ana_{n+2}=\frac{2}{a_{n+1}}+a_n for n=1,2,3n=1,2,3, ... If (a1+1a2a3)⋅(a2+1a3a4)⋅(a3+1a4a5)…(a30+1a31a32)\left(\frac{a_1+\frac{1}{a_2}}{a_3}\right) \cdot\left(\frac{a_2+\frac{1}{a_3}}{a_4}\right) \cdot\left(\frac{a_3+\frac{1}{a_4}}{a_5}\right) \ldots\left(\frac{a_{30}+\frac{1}{a_{31}}}{a_{32}}\right) =2α(61C31)=2^\alpha\left({ }^{61} C_{31}\right), then α\alpha is equal to
View answer and explanation
Correct option: C
−60-60
Option C (-60) is correct because after simplifying the recursive sequence and the product, the value of α can be determined.
675 Easy
675 of 998
Application +4 | -1
Let a1, a2, .... , a10{a_1},\,{a_2},\,....\,,\,{a_{10}} be a GP. If a3a1=25,\frac{{{a_3}}}{{{a_1}}} = 25, then a9a5\frac{{{a_9}}}{{{a_5}}} equals
View answer and explanation
Correct option: D
54{5^4}
Option D (54{5^4}) is correct because the ratio a9a5\frac{{{a_9}}}{{{a_5}}} simplifies to r4{r^4}, and given a3a1=25\frac{{{a_3}}}{{{a_1}}} = 25, we find r4=54{r^4} = {5^4}.

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
676 Easy
676 of 998
Application +4 | -1
If a,b,ca, b, c are in H.P., then
View answer and explanation
Correct option: B
an+cn>2bna^n+c^n>2 b^n
Option B (an+cn>2bna^n+c^n>2 b^n) is correct because it is derived from the relationship between arithmetic and harmonic means, where A.M. > H.M., and using the inequality an+cn2>(a+c2)n\frac{a^n+c^n}{2}>\left(\frac{a+c}{2}\right)^n.
677 Easy
677 of 998
Analyzing +4 | -1
Let a,ba, b, c be in A.P. and ∣a∣<1,∣b∣<1,∣c∣<1|a|<1,|b|<1,|c|<1. If x=1+a+a2+…x=1+a+a^2+\ldots to ∞,y=1+b+b2+…\infty, y=1+b+b^2+\ldots to ∞\infty and, z=1+c+c2+…z=1+c+c^2+\ldots to ∞\infty, then x,yx, y, zz are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because if a, b, c are in AP, then 1/(1-a), 1/(1-b), 1/(1-c) are in HP.
678 Easy
678 of 998
Analyzing +4 | -1
The minimum value of f(x)=aax+a1−axf(x) = {a^{{a^x}}} + {a^{1 - {a^x}}}, where a,  x∈Ra,\,\,x \in R and a>0a > 0, is equal to
View answer and explanation
Correct option: C
2a2\sqrt a
Option C (2a2\sqrt a) is correct because it is found by applying the Arithmetic Mean (AM) >= Geometric Mean (GM) inequality to the given expression.
679 Easy
679 of 998
Analyzing +4 | -1
If ∣x∣<1,∣y∣<1|x|<1,|y|<1 and x≠yx \neq y, then the sum to infinity of the following series:

(x+y)+(x2+xy+y2)+(x3+x2y+xy2+y3)+…(x+y)+\left(x^2+x y+y^2\right)+\left(x^3+x^2 y+x y^2+y^3\right)+\ldots, is
View answer and explanation
Correct option: B
x+y−xy(1−x)(1−y)\frac{x+y-x y}{(1-x)(1-y)}
Option B (x+y−xy(1−x)(1−y)\frac{{x+y-xy}}{{(1-x)(1-y)}}) is correct because the series can be transformed into a difference of geometric series, leading to this simplified expression.
680 Easy
680 of 998
Application +4 | -1
The first, second and middle terms of an AP are a,b,ca, b, c respectively. Their sum is
View answer and explanation
Correct option: B
2c(c−a)b−a+c\frac{2 c(c-a)}{b-a}+c
Option D (2c(c−a)b−a+c\frac{{2 c(c-a)}}{{b-a}}+c) is correct because it represents the sum of an A.P. with an odd number of terms, where the middle term is known.
681 Easy
681 of 998
Analyzing +4 | -1
If x=∑n=0∞an,y=∑n=0∞bn,z=∑n=0∞(ab)nx=\sum\limits_{n=0}^{\infty} a^n, y=\sum\limits_{n=0}^{\infty} b^n, z=\sum\limits_{n=0}^{\infty}(a b)^n, where a,b<1a, b<1, then
View answer and explanation
Correct option: B
xz+yz=xy+zx z+y z=x y+z
Option B (xz+yz=xy+zx z+y z=x y+z) is correct because it is derived from the relationships between x, y, and z, which are sums of infinite geometric series involving a and b.
682 Easy
682 of 998
Analyzing +4 | -1
If ∑r=1nar=16n(n+1)(n+2)\sum \limits_{r = 1}^n {a_r} = \frac{1}{6}n\left( {n + 1} \right)\left( {n + 2} \right) for all n≥1,n \ge 1, then lim⁡n→∞∑r=1n1ar\lim\limits_{n \to \infty } \sum \limits_{r = 1}^n \frac{1}{{{a^r}}} is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the limit of the sum converges to 2 as n approaches infinity, after simplifying the series.
683 Easy
683 of 998
Understanding +4 | -1
The sum of 1+3+5+7+⋯1+3+5+7+\cdots up to nn terms is
View answer and explanation
Correct option: C
n2n^{2}
Option C (n2n^{2}) is correct because the sum of the first n odd numbers is n2n^{2}.
684 Easy
684 of 998
Understanding +4 | -1
If a,ba, b and cc are in APA P, then which of the following will also be in AP?
View answer and explanation
Correct option: A
3+a,3+b,3+c\sqrt{3}+a, \sqrt{3}+b, \sqrt{3}+c
Option A (3+a,3+b,3+c\sqrt{3}+a, \sqrt{3}+b, \sqrt{3}+c) is correct because adding a constant to each term of an arithmetic progression results in another arithmetic progression.
685 Easy
685 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: D
96
Option D (96) is correct because the sum of the first 9 terms of the series is calculated by simplifying the general term and then summing the series.
686 Easy
686 of 998
Application +4 | -1
In the sequence 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8,…1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, \ldots, where nn consecutive terms have the value nn, the 1025th term is
View answer and explanation
Correct option: B
2102^{10}
Option B (2102^{10}) is correct because the 1025th term falls within the block of terms where the value is 2102^{10} in the given sequence.
687 Easy
687 of 998
Application +4 | -1
The sum to nn terms of the infinite series 1.32+2.52+3.72+...∞{1.3^2} + {2.5^2} + {3.7^2} + ...\infty
View answer and explanation
Correct option: A
n6(n+1)(6n2+14n+7)\frac{n}{6}\left( {n + 1} \right)\left( {6{n^2} + 14n + 7} \right)
Option A is correct because the sum to n terms of the given series is derived by expressing the general term, summing it, and simplifying the expression.
688 Easy
688 of 998
Application +4 | -1
99th term of the series 2+7+14+23+34…2 + 7 + 14 + 23 + 34 \ldots is
View answer and explanation
Correct option: A
9998
Option A (9998) is correct because the 99th term of the given series is calculated by finding the general term and substituting n=99.
689 Easy
689 of 998
Analyzing +4 | -1
The sum of the first three terms of a G.P.is SS and their product is 27. Then all such SS lie in
View answer and explanation
Correct option: A
(−∞,−3]∪[9,∞)(-\infty,-3] \cup[9, \infty)
Option A ((−∞,−3]∪[9,∞))((-\infty,-3] \cup[9, \infty)) is correct because it represents the range of possible values for the sum SS based on the properties of the geometric progression and the AM-GM inequality.
690 Easy
690 of 998
Analyzing +4 | -1
If log⁡x{log⁡4(log⁡x(5x2+4x3))}=0,{\log _x}\left\{ {{{\log }_4}\left( {{{\log }_x}\left( {5{x^2} + 4{x^3}} \right)} \right)} \right\} = 0, then
View answer and explanation
Correct option: D
5
Option D (5) is correct because solving the logarithmic equation step-by-step leads to the solution x = 5.
Revision notes and key takeaways

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

Showing 661-690 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 →