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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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391 Easy
391 of 998
Understanding +4 | -1
∑j=1n∑i=1ni\sum\limits_{{j=1}}^{n} \sum\limits_{{i=1}}^{n} i is equal to
View answer and explanation
Correct option: C
n2(n+1)2\frac{n^{2}(n+1)}{2}
Option C (n2(n+1)2\frac{{n^{{2}}(n+1)}}{{2}}) is correct because the inner sum is n(n+1)2\frac{{n(n+1)}}{{2}}, and the outer sum multiplies this by n.
392 Easy
392 of 998
Application +4 | -1
The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternativety positive and negative, then the first term is
View answer and explanation
Correct option: D
-12
Option D (-12) is correct because it satisfies the given conditions of the geometric progression with alternating signs.
393 Easy
393 of 998
Application +4 | -1
The ratio of the sum of first three terms of a G.P. to the sum of first 6 terms of the G.P. is 125:152125: 152, then common ratio of the G.P. is
View answer and explanation
Correct option: C
35\frac{3}{5}
Option C (3/5) is correct because, given the ratio of the sums, the common ratio of the G.P. is 3/5.
394 Easy
394 of 998
Analyzing +4 | -1
The sets S1,S2,S3…{S_1},{S_2},{S_3} \ldots are given by S1={21},S2={32,52},S3={43,73,103},S4={54,94,134,174},…{S_1} = \left\{ {\frac{2}{1}} \right\},{S_2} = \left\{ {\frac{3}{2},\frac{5}{2}} \right\},{S_3} = \left\{ {\frac{4}{3},\frac{7}{3},\frac{{10}}{3}} \right\},{S_4} = \left\{ {\frac{5}{4},\frac{9}{4},\frac{{13}}{4},\frac{{17}}{4}} \right\}, \ldots Then, the sum of the numbers in the set S25{S_{25}} is
View answer and explanation
Correct option: D
326
Option D (326) is correct because the sum of the arithmetic series in set S_{25} is calculated correctly using the AP sum formula.
395 Easy
395 of 998
Analyzing +4 | -1
Let the sequence a1,a2,a3,…,an,…a_1, a_2, a_3, \ldots, a_n, \ldots form an A.P. Then, a12−a22+a32−a42+…+a2n−12−a2n2a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2 n-1}^2-a_{2 n}^2 is equal to
View answer and explanation
Correct option: A
n2n−1(a12−a2n2)\frac{n}{2 n-1}\left(a_1^2-a_{2 n}^2\right)
Option A (n2n−1(a12−a2n2)\frac{n}{2 n-1}\left(a_1^2-a_{2 n}^2\right)) is correct because it is derived by simplifying the given expression using the properties of an arithmetic progression.
396 Easy
396 of 998
Application +4 | -1
Let Sn{S_n} be the sum of the first nn terms of an arithmetic progression, If S3n=3S2n{S_{3n}} = 3{S_{2n}}, then the value of S4nS2n\frac{{{S_{4n}}}}{{{S_{2n}}}} is
View answer and explanation
Correct option: A
6
Option A (6) is correct because it is the value of the ratio of the sums, calculated using the properties of arithmetic progressions.
397 Easy
397 of 998
Application +4 | -1
If 17 arithmetic means are inserted between 7/27 / 2 and −83/2-83 / 2, then the 17 th AM\mathrm{AM} is
View answer and explanation
Correct option: C
-39
Option C (-39) is correct because it is the 17th arithmetic mean inserted between 7/2 and -83/2, calculated using the common difference.
398 Easy
398 of 998
Application +4 | -1
Let a1,a2,a3……….a_{1}, a_{2}, a_{3} \ldots \ldots \ldots .. be terms on A.P. If a1+a2+………..apa1+a2+………..+aq=p2q2,p≠q\frac{a_{1}+a_{2}+\ldots \ldots \ldots . . a_{p}}{a_{1}+a_{2}+\ldots \ldots \ldots . .+a_{q}}=\frac{p^{2}}{q^{2}}, p \neq q, then a6a21\frac{a_{6}}{a_{21}} equals
View answer and explanation
Correct option: D
1141\frac{11}{41}
Option D (11/41) is correct because it is obtained by substituting appropriate values into the formula for the sum of an arithmetic progression and simplifying.
399 Easy
399 of 998
Application +4 | -1
If a1,a2,…,ana_1, a_2, \ldots, a_n are in HP\mathrm{HP}, then the expression a1a2+a2a3+…+an−1ana_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n is equal to
View answer and explanation
Correct option: D
(n−1)a1an(n-1) a_1 a_n
Option D ((n−1)a1an(n-1) a_1 a_n) is correct because the sum of the product of consecutive terms in a harmonic progression can be simplified to this expression.
400 Easy
400 of 998
Application +4 | -1
If the A.M. and G.M. of two numbers aa and bb are 2 and 4 respectively, their H.M. is
View answer and explanation
Correct option: D
8
Option D (8) is correct because it is the harmonic mean calculated using the given arithmetic and geometric means.
401 Easy
401 of 998
Analyzing +4 | -1
Let {an}n=0∞\left\{a_n\right\}_{n=0}^{\infty} be a sequence such that a0=a1=0a_0=a_1=0 and an+2=3an+1−2an+1,∀n≥0a_{n+2}=3 a_{n+1}-2 a_n+1, \forall n \geq 0
Then a25 a23−2a25 a22−2a23 a24+4a22 a24a_{25}\, a_{23}-2 a_{25}\, a_{22}-2 a_{23}\, a_{24}+4 a_{22}\, a_{24} is equal to
View answer and explanation
Correct option: B
528
Option B (528) is correct because by recursively applying the given sequence relation and simplifying the expression, the result is 528.
402 Easy
402 of 998
Application +4 | -1
If H1,H2{H_1},{H_2} are two harmonic means between two positive numbers aa and b  (a≠b)b\;\left( {a \ne b} \right), AA and GG are the arithmetic and geometric means between aa and bb, then H2+H1H2H1\frac{{{H_2} + {H_1}}}{{{H_2}{H_1}}} is
View answer and explanation
Correct option: D
2AG2\frac{{2A}}{{{G^2}}}
Option D (2AG2\frac{{2A}}{{{G^2}}}) is correct because it is derived by simplifying the expression using the definitions of harmonic, arithmetic, and geometric means.
403 Easy
403 of 998
Application +4 | -1
Let a,b∈R+a, b \in R^{+}, such that log⁡27a+log⁡9b=72\log _{27} a+\log _9 b=\frac{7}{2} and log⁡27b+log⁡9a=23\log _{27} b+\log _9 a=\frac{2}{3}, then aba b is equal to:
View answer and explanation
Correct option: B
243
Option B (243) is correct because by converting the logarithms to a common base and solving the system of equations, we find that ab=243ab = 243.
404 Easy
404 of 998
Application +4 | -1
If ∑i=121ai=693\sum\limits_{{i=1}}^{21} a_i=693, where a1,a2,…,a21a_1, a_2, \ldots, a_{21} are in APA P, then the value of ∑i=010a2r+1\sum\limits_{{i=0}}^{10} a_{2 r+1} is
View answer and explanation
Correct option: B
363
Option B, 363, is correct because using the properties of an arithmetic progression, the sum can be simplified to 5(a1+a21)+a115(a_1 + a_{21}) + a_{11}, which equals 363.
405 Easy
405 of 998
Analyzing +4 | -1
Statement -1 The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) +…+(361+380+400)+\ldots+(361+380+400) is 8000 .

Statement 2∑k=1n(k3−(k−1)3)=n32 \sum_{k=1}^n\left(k^3-(k-1)^3\right)=n^3 for any natural number nn.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
Option A (Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1) is correct because the sum of the series telescopes to 8000, which is 20 cubed, and the second statement correctly represents the telescoping sum.
406 Easy
406 of 998
Application +4 | -1
Sum of the series S=12−22+32−42+S=1^2-2^2+3^2-4^2+ …−20082+20092\ldots-2008^2+2009^2 is
View answer and explanation
Correct option: A
2019045
Option A (2019045) is correct because the sum of the series can be simplified by pairing terms and using the formula for the sum of the first n integers.
407 Easy
407 of 998
Application +4 | -1
The sum of the intercepts cut off by the axes on the lines x+y=a,x+y=ar,x+y=ar2,…x+y=a, x+y=a r, x+y=a r^2, \ldots where a≠0a \neq 0 and r=1/2r=1 / 2 is
View answer and explanation
Correct option: C
22a2 \sqrt{2} a
The correct answer is 22a2 \sqrt{{2}} a because the sum of intercepts forms a geometric series with a common ratio of 1/2, and the sum to infinity is calculated.
408 Easy
408 of 998
Analyzing +4 | -1
The sum of the series 2{7−1+3−1⋅7−3+5−1⋅7−5+…}2\left\{ {{7^{ - 1}} + {3^{ - 1}} \cdot {7^{ - 3}} + {5^{ - 1}} \cdot {7^{ - 5}} + \ldots } \right\} is
View answer and explanation
Correct option: A
log⁡e(43){\log _e}\left( {\frac{4}{3}} \right)
Option A (log⁡e(43){\log _e}(\frac{4}{3})) is correct because the series can be recognized as a Taylor series expansion of a logarithmic function.
409 Easy
409 of 998
Application +4 | -1
If the nnth terms of two series 3+10+17+…3+10+17+\ldots and 63+65+67+…63+65+67+\ldots are equal, then n=n=
View answer and explanation
Correct option: C
13
Option C (13) is correct because the nth terms of both series are equal when n = 13, as found by setting the expressions for the nth terms equal to each other and solving for n.
410 Easy
410 of 998
Analyzing +4 | -1
A GP consists an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying odd places, then the common ratio will be equal to
View answer and explanation
Correct option: C
4
Option C (4) is correct because the common ratio is 4, derived from the given condition about the sum of terms in the GP.
411 Easy
411 of 998
Analyzing +4 | -1
If x=1+log⁡abc,y=1+log⁡bca,z=1+log⁡cab,x=1+\log _a b c, y=1+\log _b c a, z=1+\log _c a b, then xyzxy+yz+zx\frac{x y z}{x y+y z+z x} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by using logarithmic properties, we can simplify the expression and find that the given fraction equals 1.
412 Easy
412 of 998
Analyzing +4 | -1
If nn is an odd integer greater than or equal to 1 , the value of n3−(n−1)3+(n−2)3−…+(−1)n−113n^3-(n-1)^3+(n-2)^3-\ldots+(-1)^{n-1} 1^3 is
View answer and explanation
Correct option: A
(n+1)2(2n−1)4\frac{(n+1)^2(2 n-1)}{4}
Option A ((n+1)2(2n−1)4\frac{{(n+1)^2(2n-1)}}{{4}}) is correct because it is the simplified form of the given alternating sum of cubes, derived by grouping terms and applying summation formulas.
413 Easy
413 of 998
Application +4 | -1
The sum of i−2−3i+4….i - 2 - 3i + 4 \ldots . upto 100 terms, where i=−1i = \sqrt { - 1} is
View answer and explanation
Correct option: A
50(1−i)50\left( {1 - i} \right)
Option A (50(1−i)50\left( {1 - i} \right)) is correct because it is the result of summing the series by manipulating and simplifying the expression using properties of complex numbers.
414 Easy
414 of 998
Application +4 | -1
Suppose a1,a2,…a_1, a_2, \ldots are in A.P. If a8:a5=a_8: a_5= 3:23: 2, then a17:a23a_{17}: a_{23} is:
View answer and explanation
Correct option: B
3:43: 4
Option B (3:4) is correct because, given the ratio of the 8th and 5th terms, we can find the relationship between the first term and common difference, then find the ratio of the 17th and 23rd terms.
415 Easy
415 of 998
Application +4 | -1
A species has an initial population 4104^{10}. At the end of first day, the population increases by 50%50 \%. At the end of second day, it decreases by the same percentage. If the process continues in the same pattern, the number of days for the population to reach 3103^{10} is
View answer and explanation
Correct option: B
20
Option B (20) is correct because after 20 days, with alternating 50% increases and decreases, the population reaches 310.
416 Easy
416 of 998
Application +4 | -1
If AM of two terms is 9 and HM is 36 , then GMG M will be
View answer and explanation
Correct option: A
18
Option A (18) is correct because the Geometric Mean (GM) is the square root of the product of the Arithmetic Mean (AM) and Harmonic Mean (HM).
417 Easy
417 of 998
Application +4 | -1
If the ratio of sum of nn terms of two A.P's is (3n+8):(7n+15)(3 n+8):(7 n+15), then the ratio of 12ti 12^{\text {ti }} terms is-
View answer and explanation
Correct option: B
7:167: 16
Option B (7: 16) is correct because it is the ratio of the 12th terms obtained by substituting the appropriate value of n in the given ratio of sums.
418 Easy
418 of 998
Application +4 | -1
The sum of the series 1⋅2⋅3+2⋅3⋅4+3⋅4⋅5+...1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + ... to nn terms is
View answer and explanation
Correct option: C
14n(n+1)(n+2)(n+3)\frac{1}{4}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)
Option C (14n(n+1)(n+2)(n+3)\frac{1}{4}n(n + 1)(n + 2)(n + 3)) is correct because it represents the sum of the series 1⋅2⋅3+2⋅3⋅4+3⋅4⋅5+...1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + ... to nn terms, which can be found by expressing the general term as n(n+1)(n+2)n(n+1)(n+2) and summing it.
419 Easy
419 of 998
Application +4 | -1
If 112+122+132+⋯\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots upto ∞=π26\infty=\frac{\pi^2}{6}, then value of 112+132+152+⋯\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots upto ∞\infty is
View answer and explanation
Correct option: C
π28\frac{\pi^2}{8}
Option C (π28\frac{\pi^2}{8}) is correct because it is derived by subtracting the sum of even-numbered terms from the total sum of the series.
420 Easy
420 of 998
Application +4 | -1
The AM, HM and GM between two numbers are 14415,  15\frac{{144}}{{15}},\;15 and 12, but not necessarily in this order. Then, HM, GM and AM respectively are
View answer and explanation
Correct option: B
14415,12,  15\frac{{144}}{{15}},12,\;15
Option A is correct because the GM is always between the AM and HM. Given AM and GM, HM can be calculated and matched with the given values.
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