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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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Showing 1-30 of 564 questions
1 Medium
1 of 564
Application +4 | -1
Given the sets A={1,2,3},B={3,4},C={4,5,6}A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}, then find the following
(i) A∪(B∩C){A} \cup({B} \cap {C})
(ii) A−(B∩C)A-(B \cap C)
(iii) (B∪C)−A(B \cup C)-A
View answer and explanation
Correct option: A
(i) {1,2,3,4}\{1,2,3,4\}, (ii) {1,2,3}\{1,2,3\}, (iii) {4,5,6}\{4,5,6\}
Option A ((i) {1,2,3,4}, (ii) {1,2,3}, (iii) {4,5,6}) is correct because it accurately computes the set operations: union, intersection, and set difference.
2 Medium
2 of 564
Analyzing +4 | -1
If log⁡(5ca),log⁡(3b5c)\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right) and log⁡(a3b)\log \left(\frac{a}{3 b}\right) are in APA P, where a,ba, b and c are in GPG P, then a,b,ca, b, c are the lengths of sides of
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the given conditions lead to a situation where a triangle cannot be formed, as the sum of two sides is less than the third side.
3 Medium
3 of 564
Application +4 | -1
In an infinite G.P., if the sum of all the terms is 7 and the sum of cubes of all the terms is 49, then the common ratio between its consecutive terms is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{{1}}{{2}}) is correct because solving the equations derived from the sum of the infinite GP and the sum of the cubes of its terms leads to this common ratio.
4 Medium
4 of 564
Analyzing +4 | -1
If a,ba, b and cc are in AP and a,mb,ca, m b, c are in GP, then a,m2ba, m^2 b and cc are in
View answer and explanation
Correct option: A
HP\mathrm{HP}
Option A (HP) is correct because by using the properties of AP and GP, we can show that a, m2bm^2b, and c are in HP.
5 Medium
5 of 564
Analyzing +4 | -1
If ∑r=1nr4=f(n)\sum_{r=1}^n r^4=f(n), then ∑r=1n(2r−1)4\sum_{r=1}^n(2 r-1)^4 is equal to
View answer and explanation
Correct option: A
f(2n)−16f(n)f(2 n)-16 f(n)
Option A (f(2n)−16f(n)f(2 n)-16 f(n)) is correct because the sum of (2r-1)4 from r=1 to n can be expressed as the difference between the sum of r4 from r=1 to 2n and 16 times the sum of r4 from r=1 to n.
6 Medium
6 of 564
Application +4 | -1
Let S1,S2,…S_1, S_2, \ldots be squares such that for each n≥1n \geq 1, the length of a side of SnS_n equals the length of the diagonal of Sn+1S_{n+1}. If the length of a side of S1S_1 is 10 cm10 \mathrm{~cm} and the area of SnS_n less than 1sqcm1 \mathrm{sq} \mathrm{cm}, then the least value of nn is
View answer and explanation
Correct option: B
8
Option B (8) is correct because the area of the nth square must be less than 1, which leads to the inequality 2n−1>1002^{n-1} > 100, and the smallest integer n that satisfies this is 8.
7 Medium
7 of 564
Application +4 | -1
The fifth and 31st terms of an AP are, respectively, 1 and -77. If kk th term of the given AP is -17 , then kk is
View answer and explanation
Correct option: C
11
Option C (11) is correct because solving the arithmetic progression equations leads to k = 11.
8 Medium
8 of 564
Analyzing +4 | -1
Let ana_n be the nnth term of a GP of positive numbers.
If ∑n=1100a2n=α\sum_{n=1}^{100} a_{2 n}=\alpha and ∑n=1100a2n−1=β\sum_{n=1}^{100} a_{2 n-1}=\beta, such that α≠β\alpha \neq \beta, then the common ratio is
View answer and explanation
Correct option: A
αβ\frac{\alpha}{\beta}
Option A (αβ\frac{\alpha}{\beta}) is correct because the ratio of the sums of even and odd terms in a GP directly gives the common ratio.
9 Medium
9 of 564
Application +4 | -1
If x,y,zx, y, z are three real numbers of the same sign then the value of xy+yz+zx\frac{x}{y}+\frac{y}{z}+\frac{z}{x} lies in the interval
View answer and explanation
Correct option: B
[3,∞)[3, \infty)
Option B ([3,∞)[3, \infty)) is correct because, by applying the AM-GM inequality, the value of the expression is always greater than or equal to 3.
10 Medium
10 of 564
Application +4 | -1
If log⁡102,log⁡10(2x−1),log⁡10(2x+3)\log _{10} 2, \log _{10}\left(2^x-1\right), \log _{10}\left(2^x+3\right) are in AP\mathrm{AP}, then
View answer and explanation
Correct option: C
x=log⁡25x=\log _2 5
Option C (x=log⁡25x=\log _2 5) is correct because it is derived by using the AP property and solving the resulting exponential equation.
11 Medium
11 of 564
Application +4 | -1
Let S1,S2,S3S_1, S_2, S_3 be the sums of nn terms of three series in A.P., the first term of each being 1 and the common differences 1,2,31,2,3, resp. If S1+S3=λS2S_1+S_3=\lambda S_2 then the value of λ\lambda is:
View answer and explanation
Correct option: B
2
Option B (2) is correct because substituting the given series into the equation S1+S3=λS2S_1 + S_3 = \lambda S_2 and solving for λ yields 2.
12 Medium
12 of 564
Application +4 | -1
If (2n+r)r,n∈N,r∈N(2 n+r) r, n \in N, r \in N is expressed the sum of kk consecutive odd natural numbers, then kk is equal to
View answer and explanation
Correct option: A
rr
Option A (rr) is correct because (2n+r)r(2n+r)r can be expressed as the difference of two squares, representing the sum of rr consecutive odd natural numbers.
13 Medium
13 of 564
Application +4 | -1
If x,∣x+1∣,∣x−1∣x,\left| {x + 1} \right|,\left| {x - 1} \right| are first three terms of an A.P., then the sum of its first 20 terms is
View answer and explanation
Correct option: B
350,180350,180
Option B (350, 180) is correct because it provides the two possible sums of the first 20 terms of the A.P., calculated based on the two possible values of x.
14 Medium
14 of 564
Analyzing +4 | -1
The sum to infinite term of the series 1+23+632+1033+1434+⋯1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\frac{14}{3^{4}}+\cdots is
View answer and explanation
Correct option: A
3
Option A (3) is correct because by using the method of difference and summing the resulting geometric series, the sum to infinity converges to 3.
15 Medium
15 of 564
Analyzing +4 | -1
Define an=n2+20a_n=n^2+20 for n=1,2,3,…n=1,2,3, \ldots. Let dnd_n be the greatest common divisor of ana_n and an+1a_{n+1} and dd be the least common multiple of dnd_n for n=1,2,3,…n=1,2,3, \ldots. Then the infinite sum of the series
d+d4+d42+d43+⋯d+\frac{d}{4}+\frac{d}{4^2}+\frac{d}{4^3}+\cdots
is
View answer and explanation
Correct option: C
108
Option C (108) is correct because it is the sum of the infinite geometric series with first term d=81 and common ratio 1/4.

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16 Medium
16 of 564
Application +4 | -1
Sum of the series (1+x)+(1+x+x2)+(1+x+x2+x3)+…n(1+x)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\ldots n terms is
View answer and explanation
Correct option: A
11−x[n−x2(1−xn)1−x]\frac{1}{1-x}\left[n-\frac{x^2\left(1-x^n\right)}{1-x}\right]
Option C (11−x[n−x(1−xn)1−x]\frac{1}{1-x}\left[n-\frac{x(1-x^n)}{1-x}\right]) is correct because the sum of the series can be found by expressing each term as a geometric series and then summing the resulting series.
17 Medium
17 of 564
Understanding +4 | -1
If a1,  a2,  a3...,  a20{a_1},\;{a_2},\;{a_3}...,\;{a_{20}} are AM’s between 13 and 67, then the maximum value of a1,  a2,  a3...,  a20{a_1},\;{a_2},\;{a_3}...,\;{a_{20}} is equal to
View answer and explanation
Correct option: B
(40)20{\left( {40} \right)^{20}}
Option B ((40)20{\left( {40} \right)^{20}}) is correct because the maximum value of the product occurs when all the arithmetic means are equal to their average.
18 Medium
18 of 564
Understanding +4 | -1
Two sequences <an>< {a_n} > and <bn>< {b_n} > are defined by an=log⁡(5n+13n−1),bn={log⁡(53)}n,{a_n} = \log \left( {\frac{{{5^{n + 1}}}}{{{3^{n - 1}}}}} \right),{b_n} = {\left\{ {\log \left( {\frac{5}{3}} \right)} \right\}^n}, then
View answer and explanation
Correct option: A
<an>< {a_n} > is an A.P. and <bn>< {b_n} > is a G.P.
Option A is correct because the sequence a_n forms an arithmetic progression, while b_n forms a geometric progression based on their logarithmic properties.
19 Medium
19 of 564
Application +4 | -1
The least positive integer n\mathrm{n} such that 1−23−232−….−23n−1<11001-\frac{2}{3}-\frac{2}{3^{2}}-\ldots .-\frac{2}{3^{\mathrm{n}-1}}<\frac{1}{100}, is:
View answer and explanation
Correct option: B
5
Option B (5) is correct because it is the least positive integer n satisfying the given inequality after simplifying the geometric series and solving for n.
20 Medium
20 of 564
Application +4 | -1
The coefficient of xn{x^n} in the series 1+a+bx1!+(a+bx)22!+(a+bx)33!=…∞1 + \frac{{a + bx}}{{1!}} + \frac{{{{\left( {a + bx} \right)}^2}}}{{2!}} + \frac{{{{\left( {a + bx} \right)}^3}}}{{3!}} = \ldots \infty is
View answer and explanation
Correct option: C
ea.bnn!{e^a}.\frac{{{b^n}}}{{n!}}
Option C (ea.bnn!{e^a}.\frac{{{b^n}}}{{n!}}) is correct because the series represents the exponential function ea+bxe^{a+bx}, and the coefficient of xnx^n can be found by expanding ebxe^{bx} and multiplying by eae^a.
21 Medium
21 of 564
Application +4 | -1
If a1,    a2,    a3,…,  a24{a_1},\;\;{a_2},\;\;{a_3}, \ldots ,\;{a_{24}} are in arithmetic progression and a1+a5+a10+a15+a20+a24=225,{a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225, then
a1+a2+a3+...+a23+a24{a_1} + {a_2} + {a_3} + ... + {a_{23}} + {a_{24}} is equal to
View answer and explanation
Correct option: D
900
Option D (900) is correct because, in an AP, the sum of terms equidistant from the beginning and end is constant, allowing us to find the sum of all 24 terms.
22 Medium
22 of 564
Application +4 | -1
If log⁡4(3x2+11x)>1,{\log _4}\left( {3{x^2} + 11x} \right) > 1, then xx lies in the interval
View answer and explanation
Correct option: D
(−∞,−4)∪(1/3  ∞)\left( { - \infty , - 4} \right) \cup \left( {1/3\;\infty } \right)
Option D is correct because solving the logarithmic inequality and considering the domain of the logarithm leads to the interval (−∞,−4)∪(1/3,∞)\left( { - \infty , - 4} \right) \cup \left( {1/3,\infty } \right).
23 Medium
23 of 564
Application +4 | -1
log⁡e1+3x1−2x{\log _e}\frac{{1 + 3x}}{{1 - 2x}} is equal to
View answer and explanation
Correct option: C
5x−5x22+35x33−  …5x - \frac{{5{x^2}}}{2} + \frac{{35{x^3}}}{3} - \; \ldots
Option D (5x+5x22+35x33+…5x + \frac{{5{x^2}}}{2} + \frac{{35{x^3}}}{3} + \ldots) is correct because expanding the logarithmic expression using the series expansion of log(1+x) and log(1-x) results in the given series.
24 Medium
24 of 564
Application +4 | -1
If <an>< a_n > is an A.P., then D=∣apaqarpqr111∣=\mathrm{D}=\left|\begin{array}{ccc}a_p & a_q & a_r \\ p & q & r \\ 1 & 1 & 1\end{array}\right|=
View answer and explanation
Correct option: A
0
Option A (0) is correct because the determinant is zero due to the properties of arithmetic progressions and determinants.
25 Medium
25 of 564
Analyzing +4 | -1
STATEMENT-1: If a1,a2,a3,_____a16{a_1},{a_2},{a_3},\_\_\_\_\_{a_{16}} is an A.P such that a1+a4+a7+−−a16=147{a_1} + {a_4} + {a_7} + - - {a_{16}} = 147, then a1+a6+a11+a16=98{a_1} + {a_6} + {a_{11}} + {a_{16}} = 98
STATEMENT-2: In an A.P the sum of the terms equidistant from the beginning and the end is always same and it is equal to the sum of first and last term
View answer and explanation
Correct option: A
Statement 1 is true statement 2 is true, statement 2 is a correct explanation of statement 1
Statement 1 is true, statement 2 is true, and statement 2 is a correct explanation of statement 1 because the sum of terms equidistant from the beginning and end is constant in an A.P.
26 Medium
26 of 564
Application +4 | -1
If a,b,c,d,ea, b, c, d, e and ff are in AP, then e=ce=c is equal to
View answer and explanation
Correct option: B
2(d−c)2(d-c)
Option B (2(d−c)2(d-c)) is correct because in an arithmetic progression, the difference between consecutive terms is constant, so e−ce-c can be expressed as 2(d−c)2(d-c).
27 Medium
27 of 564
Application +4 | -1
If the sum to infinity of the series 1+4x+7x2+10x3+………1+4 x+7 x^2+10 x^3+\ldots \ldots \ldots is 3516\frac{35}{16}, where ∣x∣<1|x|<1 then x{x} is equal to
View answer and explanation
Correct option: B
15\frac{1}{5}
Option B (15\frac{1}{5}) is correct because it solves for x using the sum to infinity formula for an arithmetico-geometric series.
28 Medium
28 of 564
Application +4 | -1
Sum of the scries 9+99+999+…9+99+999+\ldots upto nn terms is
View answer and explanation
Correct option: B
19(10n+1−9n−10)\frac{1}{9}\left(10^{n+1}-9 n-10\right)
Option B is the correct answer because the sum of the series can be expressed and simplified to 19(10n+1−9n−10)\frac{1}{9}(10^{n+1}-9 n-10).
29 Medium
29 of 564
Application +4 | -1
The arithmetic mean and the geometric mean of two numbers are 9 and 4 respectively and also the two numbers are roots of a quadratic equation. Then, the quadratic equation is ______ (k≠0)(k \neq 0)
View answer and explanation
Correct option: D
k(x2−18x+16)=0k\left(x^2-18 x+16\right)=0
Option D (k(x2−18x+16)=0k(x^2-18x+16)=0) is correct because the sum of the roots is 18 and the product is 16, which corresponds to the quadratic equation x2−(sum of roots)x+(product of roots)=0x^2 - (sum \ of \ roots)x + (product \ of \ roots) = 0.
30 Medium
30 of 564
Application +4 | -1
Let A1,A2A_1, A_2 be two arithmetic means, G1G_1, G2G_2 be two geometric means, and H1,H2H_1, H_2 be two harmonic means between two positive numbers aa and bb, then values G1G2H1H2⋅H1+H2A1+A2\frac{G_1 G_2}{H_1 H_2} \cdot \frac{H_1+H_2}{A_1+A_2} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because after simplifying the given expression using the properties of arithmetic, geometric, and harmonic means, the result equals 1.
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