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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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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 511-540 of 998 questions

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511 Easy
511 of 998
Application +4 | -1
If 12+22+32+…+20092=(2009)(4019)(335)1^2+2^2+3^2+\ldots+2009^2= (2009) (4019) (335) and

(1)(2009)+(2)(2008)+(3)(2007)+…+(2009)(1)=(2009)(335)(x)(1) (2009) + (2) (2008) + (3) (2007) +\ldots+(2009)(1)=(2009)(335)(x),

then xx equals
View answer and explanation
Correct option: A
2011
Option A (2011) is correct because by adding the two given series and using the formula for the sum of squares and the sum of integers, we can solve for x.
512 Easy
512 of 998
Application +4 | -1
Number of value(s) of 'xx' satisfying the equation log⁡2(log⁡3(x2))=1\log _2\left(\log _3\left(x^2\right)\right)=1 is/are:
View answer and explanation
Correct option: C
2
Option C is correct because solving the logarithmic equation leads to two solutions: x = 3 and x = -3.
513 Easy
513 of 998
Application +4 | -1
If a,b,ca,b,c are in H.P., then the value of b+ab−a+b+cb−c\frac{{b + a}}{{b - a}} + \frac{{b + c}}{{b - c}} is
View answer and explanation
Correct option: B
2
Option B (2) is correct because using the property of Harmonic Progression and applying componendo dividendo simplifies the expression to 2.
514 Easy
514 of 998
Application +4 | -1
If sin⁡θ,2(sin⁡θ+1),6sin⁡θ+6\sin \theta, \sqrt{2}(\sin \theta+1), 6 \sin \theta+6 are in GP\mathrm{GP}, then the fifth term is
View answer and explanation
Correct option: C
162
Option C (162) is correct because by finding the common ratio and first term, we can calculate the fifth term of the GP.
515 Medium
515 of 998
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.
516 Easy
516 of 998
Application +4 | -1
The sum of 11 terms of an A.P. whose middle term is 30, is
View answer and explanation
Correct option: B
330
Option B (330) is correct because the sum of an arithmetic progression is equal to the number of terms multiplied by the middle term.
517 Medium
517 of 998
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.
518 Easy
518 of 998
Application +4 | -1
Let f(n)=[15+3n100]nf(n)=\left[\frac{1}{5}+\frac{3 n}{100}\right] n, where [x][x] denotes the greatest integer less than or equal to xx. Then
∑n=161f(n)\sum\limits_{{n=1}}^{61} f(n)
View answer and explanation
Correct option: D
1661
Option D (1661) is correct because by evaluating the function for different ranges of n, we can simplify the summation and calculate the result.
519 Easy
519 of 998
Application +4 | -1
If 12+22+…+n2=10151^2+2^2+\ldots+n^2=1015, then value of nn is
View answer and explanation
Correct option: B
14
Option B (14) is correct. By using the formula for the sum of squares and solving the resulting equation, we find n = 14.
520 Medium
520 of 998
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.
521 Easy
521 of 998
Understanding +4 | -1
If {an}\left\{ {{a_n}} \right\} is a sequence with a0=p{a_0} = p and an−an−1=ran−1{a_n} - {a_{n - 1}} = r{a_{n - 1}} for n≥1n \ge 1, then the terms of the sequence are in
View answer and explanation
Correct option: B
A geometric progression
Option B (A geometric progression) is correct because the recursive relation implies a constant ratio between consecutive terms.
522 Medium
522 of 998
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.
523 Easy
523 of 998
Application +4 | -1
For a sequence <an>,a1=5< a_n > , a_1=5, and ar+1ar=12 ∀ r\frac{a_{r+1}}{a_r}=\frac{1}{2} \, \forall \, r, then ∑n=1∞a2n−1\sum\limits_{{n=1}}^{\infty} a_{2 n-1} is
View answer and explanation
Correct option: C
20/320 / 3
Option C (20/320 / 3) is correct because the sum of the specified infinite geometric series is calculated correctly.
524 Medium
524 of 998
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.
525 Easy
525 of 998
Application +4 | -1
nn arithmetic means are inserted between 3 and 17. If the ratio of first to last mean is 1 : 3, then nn is
equal to:
View answer and explanation
Correct option: C
6
Option C (6) is correct because by using the arithmetic means formula and the given ratio, we can solve for the number of arithmetic means.
526 Medium
526 of 998
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.
527 Easy
527 of 998
Application +4 | -1
150 workers were engaged to finish a piece of work in a certain number of days. 4 workers dropped the second day, 4 more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is
View answer and explanation
Correct option: C
25
Option C (25) is correct because it is the number of days calculated by setting up an AP series representing the decreasing number of workers and solving for the original number of days.
528 Medium
528 of 998
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.
529 Medium
529 of 998
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.
530 Medium
530 of 998
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.
531 Medium
531 of 998
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.
532 Easy
532 of 998
Analyzing +4 | -1
The sum of the series 12⋅  221  !+22⋅  322  !+32⋅  423  !+…,\frac{{{1^2} \cdot \;{2^2}}}{{1\;!}} + \frac{{{2^2} \cdot \;{3^2}}}{{2\;!}} + \frac{{{3^2} \cdot \;{4^2}}}{{3\;!}} + \ldots , is
View answer and explanation
Correct option: A
27e27e
Option A (27e27e) is correct because the series can be decomposed into sums involving ee, resulting in 27e27e.
533 Medium
533 of 998
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.
534 Easy
534 of 998
Application +4 | -1
Solution of 0<∣3x+1∣<130 < |3 x+1| < \frac{1}{3} is
View answer and explanation
Correct option: C
(−4/9,−2/9)−{−1/3}(-4 / 9,-2 / 9)-\{-1 / 3\}
Option C ((-4 / 9,-2 / 9)-{-1 / 3}) is correct because it represents the interval where the absolute value condition is met, excluding the point where the absolute value is zero.
535 Easy
535 of 998
Analyzing +4 | -1
∑k=1∞1k!(∑n=1∞2n−1)\sum \limits_{k = 1}^\infty \frac{1}{{k!}}\left( {\sum \limits_{n = 1}^\infty {2^{n - 1}}} \right) is equal to
View answer and explanation
Correct option: D
e2−e{e^2} - e
Option D (e2−e{e^2} - e) is correct because it is derived by splitting the sum into two separate exponential series and evaluating them.
536 Easy
536 of 998
Application +4 | -1
If 2(y−a)2\left( {y - a} \right) is the H.M. between y−xy - x and y−z,y - z, then x−a,  y−a,  z−ax - a,\;y - a,\;z - aare in
View answer and explanation
Correct option: B
G.P.
The correct answer is B (G.P.) because the given harmonic mean condition leads to the conclusion that x-a, y-a, and z-a are in geometric progression.
537 Easy
537 of 998
Application +4 | -1
The second term of an AP is (x−y)(x-y) and the fifth term is (x+y)(x+y), then its first term is
View answer and explanation
Correct option: D
x−53yx-\frac{5}{3} y
Option D (x−53yx-\frac{5}{3} y) is correct because by using the information about the second and fifth terms of the AP, we can find the first term using the properties of arithmetic progressions.
538 Easy
538 of 998
Application +4 | -1
If SnS_n denotes the sum of first nn term of an A.P., then S3n−Sn−1S2n−S2n−1\frac{S_{3 n}-S_{n-1}}{S_{2 n}-S_{2 n-1}} is equal to
View answer and explanation
Correct option: B
2n+12 n+1
Option B (2n+12 n+1) is correct because by substituting the formula for the sum of an A.P., the expression simplifies to 2n+12n+1.
539 Easy
539 of 998
Application +4 | -1
In an A.P., the first term is 1 and sum of the first pp terms is 0, then sum of the first (p+q)(p+q) terms is
View answer and explanation
Correct option: A
q(p+q)1−p\frac{q(p+q)}{1-p}
Option A (q(p+q)1−p\frac{q(p+q)}{1-p}) is correct because it is the result of calculating the sum of the first (p+q) terms using the given information about the A.P.
540 Easy
540 of 998
Application +4 | -1
Suppose a1,a2,…an,…a_1, a_2, \ldots a_n, \ldots be an arithmetic progression of natural numbers. If the ration of the sum of first five terms to the sum of first nine terms of the progression is 5:175: 17 and 110 < a15 < 120110\, < \,{a_{15}}\, < \,120, then the sum of the first ten terms of the progression is equal to
View answer and explanation
Correct option: B
380
Option B is correct because by using the given ratio of sums and the condition on a_15, we can find the values of a_1 and d, and then calculate the sum of the first ten terms.
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