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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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721 Easy
721 of 998
Application +4 | -1
Let a1,a2,…an,…a_1, a_2, \ldots a_n, \ldots be a G.P. of positive terms such that ∑n=1100a2n+1=200,∑n=1100a2n=100\sum\limits_{n=1}^{100} a_{2 n+1}=200, \sum\limits_{n=1}^{100} a_{2 n}=100. Then, ∑n=1200an\sum\limits_{n=1}^{200} a_n is equal to
View answer and explanation
Correct option: B
150
Option B is correct because by using the given sums and the properties of a geometric progression, the sum of the first 200 terms can be found to be 150.
722 Easy
722 of 998
Application +4 | -1
If a1,a2,a3,a4,a5a_1, a_2, a_3, a_4, a_5 are in HP\mathrm{HP}, then a1a2+a2a3+a3a4+a4a5a_1 a_2+a_2 a_3+a_3 a_4+a_4 a_5 is equal to
View answer and explanation
Correct option: C
4a1a54 a_1 a_5
Option A (n2n^2) is correct because it represents the sum to infinity of the given series, derived using the formula for the sum of an infinite geometric progression.
723 Medium
723 of 998
Application +4 | -1
Three positive numbers form an increasing GP. If the middle terms in this GP is doubled, the new numbers are in AP. Then, the common ratio of the GP is
View answer and explanation
Correct option: B
2+32+\sqrt{3}
Option B (2+32+\sqrt{3}) is correct because it is the common ratio of the increasing GP, derived from the condition that doubling the middle term results in an AP.
724 Easy
724 of 998
Analyzing +4 | -1
If GG is the GM of the product of rr set of observation with geometric means G1,  G2,…Gr{G_1},\;{G_2}, \ldots {G_r} respectively, then GG is equal to
View answer and explanation
Correct option: B
G1,G2,….,Gr{G_1},{G_2}, \ldots .,{G_r}
Option D (G1,G2,….,Gr{G_1},{G_2}, \ldots .,{G_r}) is incorrect. Option A (log⁡G1+log⁡G2+...+log⁡Gn\log {G_1} + \log {G_2} + ... + \log {G_n}) is incorrect. Option C (log⁡G1,log⁡G2,….,log⁡Gn\log {G_1},\log {G_2}, \ldots .,\log {G_n}) is incorrect. Option B (G1,G2,….,Gr{G_1},{G_2}, \ldots .,{G_r}) is correct because the geometric mean of the product of rr sets of observations is the product of the individual geometric means.
725 Easy
725 of 998
Application +4 | -1
If a,b,ca,b,c are in H.P., then 1b−a+1b−c=\frac{1}{{b - a}} + \frac{1}{{b - c}} =
View answer and explanation
Correct option: B
1a+1c\frac{1}{a} + \frac{1}{c}
Option B, 1a+1c\frac{1}{a} + \frac{1}{c}, is correct because it represents the simplified form of the expression given that a, b, and c are in harmonic progression.
726 Medium
726 of 998
Analyzing +4 | -1
If the sum of nn terms of the series

1+45+752+1053+…1 + \frac{4}{5} + \frac{7}{{{5^2}}} + \frac{{10}}{{{5^3}}} + \ldots is l+1516(1−15n−1)−(3n−2)4(5n−1)l + \frac{{15}}{{16}}\left( {1 - \frac{1}{{{5^{n - 1}}}}} \right) - \frac{{\left( {3n - 2} \right)}}{{4\left( {{5^{n - 1}}} \right)}}, then ll is
View answer and explanation
Correct option: B
54\frac{5}{4}
Option B (54\frac{5}{4}) is correct because it is the constant term obtained after applying the formula for the sum of an arithmetico-geometric series and comparing coefficients.
727 Easy
727 of 998
Application +4 | -1
The sum of first 26 terms of an A.P. a1,a2,a3,…a_1, a_2, a_3, \ldots if a2+a6+a9+a18+a21+a25=165a_2+a_6+a_9+a_{18}+a_{21}+a_{25}=165, is
View answer and explanation
Correct option: B
715
Option B (715) is correct because the sum of the first 26 terms of an A.P. can be found using the given relationship between the terms.
728 Easy
728 of 998
Application +4 | -1
If sum of four numbers in A.P. is 28 and product of two middle terms is 45, then product of the first and last terms is
View answer and explanation
Correct option: B
13
Option B (13) is correct because by setting up equations based on the given sum and product, we can solve for the product of the first and last terms.
729 Easy
729 of 998
Application +4 | -1
If log⁡4,log⁡(2x+2),log⁡(2x+2+1)\log 4, \log \left(2^x+2\right), \log \left(2^{x+2}+1\right) are 3 consecutive terms of an arithmetic progression then xx lies between 2 consecutive natural numbers pp and q(p<q)q(p < q) find the value of (p+q)(p+q).
View answer and explanation
Correct option: A
7
The correct answer is A (7) because solving the arithmetic progression condition leads to 2x=122^x = 12, which implies xx lies between 3 and 4, hence p+q=7p+q = 7.
730 Medium
730 of 998
Analyzing +4 | -1
The sum of the series 8!+9!2!+10!3!+11!4!+…8 !+\frac{9 !}{2 !}+\frac{10 !}{3 !}+\frac{11 !}{4 !}+\ldots. up to 20 terms
View answer and explanation
Correct option: C
7!(28C20−1)7 !\left({ }^{28} C_{20}-1\right)
Option C (7!(28C20−1)7 !\left({ }^{28} C_{20}-1\right)) is correct because the sum of the series can be simplified using combinatorial identities and adding/subtracting a constant.
731 Easy
731 of 998
Analyzing +4 | -1
If log⁡yx=log⁡zy=log⁡xz,{\log _y}x = {\log _z}y = {\log _x}z, then
View answer and explanation
Correct option: D
x=y=zx = y = z
Option D (x=y=zx = y = z) is correct because the given logarithmic condition implies that x, y, and z are equal.
732 Easy
732 of 998
Understanding +4 | -1
If arithmetic mean of two positive numbers is AA, their geometric mean is GG and harmonic mean is HH, then HH is equal to
View answer and explanation
Correct option: A
G2/A{G^2}/A
Option A is correct because the harmonic mean (H) can be expressed as the square of the geometric mean (G) divided by the arithmetic mean (A).
733 Medium
733 of 998
Application +4 | -1
The maximum sum of an AP 40,38,36,34,…40,38,36,34, \ldots is
View answer and explanation
Correct option: B
420
Option B (420) is correct because it represents the maximum sum of the arithmetic progression before the terms become negative.
734 Easy
734 of 998
Application +4 | -1
If the first and the nth {n}^{\text {th }} term of aa G.P. are aa and bb respectively and PP is the product of the first nn terms, then P2=P^2=
View answer and explanation
Correct option: B
(ab)n(a b)^n
Option B ((ab)n(ab)^n) is correct because it is the square of the product of the first n terms of the G.P.
735 Easy
735 of 998
Application +4 | -1
If the first, second and last terms of an arithmetic series are a,ba,b and ccrespectively, then the number of terms is
View answer and explanation
Correct option: A
b+c−2ab−a\frac{{b + c - 2a}}{{b - a}}
Option A (b+c−2ab−a\frac{{b + c - 2a}}{{b - a}}) is correct because, using the formula for the last term of an arithmetic series and solving for the number of terms (n), we get this expression.
736 Easy
736 of 998
Application +4 | -1
Let the sum of the first nn terms of a non-constant AP a1, a2, a3....  {a_1},\,{a_2},\,{a_3}....\,\, be 50n+n(n−7)2A, 50n + \frac{{n(n - 7)}}{2}A,\, where AA is a constant. If dd is the common difference of this AP, then the ordered pair (d, a50)(d,\,{a_{50}}) is equal to
View answer and explanation
Correct option: A
(A, 50+46A)(A,\,50 + 46A)
Option A ((A, 50+46A)(A,\,50 + 46A)) is correct because after calculating the common difference and the 50th term, the ordered pair matches this option.
737 Medium
737 of 998
Analyzing +4 | -1
If log⁡xa2+ab+b2=logyb2+bc+c2=log⁡zc2+ca+a2,\frac{{\log x}}{{{a^2} + ab + {b^2}}} = \frac{{\mathrm{log}y}}{{{b^2} + bc + {c^2}}} = \frac{{\log z}}{{{c^2} + ca + {a^2}}}, then xa−b.yb−c.zc−a={x^{a - b}}.{y^{b - c}}.{z^{c - a}} =
View answer and explanation
Correct option: C
1
Option C (1) is correct because when the given ratios are manipulated and combined using logarithm properties, the result simplifies to 1.
738 Easy
738 of 998
Application +4 | -1
The coefficient of x3{x^3} in the expansion of 3x{3^x} is
View answer and explanation
Correct option: D
(log⁡3)36\frac{{{{\left( {\log 3} \right)}^3}}}{6}
Option D ((log⁡3)36\frac{{{{\left( {\log 3} \right)}^3}}}{6}) is correct because it is the coefficient of x3{x^3} in the Maclaurin series expansion of 3x{3^x}.
739 Easy
739 of 998
Application +4 | -1
In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of AP be equal to the middle term of the GP, then the middle term of the entire sequence is
View answer and explanation
Correct option: A
−1031- \frac{{10}}{{31}}
Option A (-10/31) is correct because it is the middle term calculated using the properties of AP and GP and the given conditions.
740 Easy
740 of 998
Application +4 | -1
If every even term of a series is a times the term before it and every odd term is cc times the term before it, the first term being unity, then the sum to 2n2 n terms is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the sum of the given series to 2n terms, calculated based on the provided conditions, does not match any of the given options.
741 Easy
741 of 998
Application +4 | -1
Let ana_n be the nth term an A.P. If ∑r=1100a2r=α\sum\limits_{r=1}^{100} a_{2 r}=\alpha and ∑r=1100a2r−1=β\sum\limits_{r=1}^{100} a_{2 r-1}=\beta, then the common difference of the A.P., is
View answer and explanation
Correct option: A
α−β100\frac{\alpha-\beta}{100}
Option A (α−β100\frac{\alpha-\beta}{100}) is correct because it determines the common difference of the A.P. based on the given sums of even and odd terms.
742 Hard
742 of 998
Analyzing +4 | -1
Suppose a,  b,  ca,\;b,\;c are in AP and a2,  b2,  c2{a^2},\;{b^2},\;{c^2} are in GP. If a<b<ca < b < c and a+b+c=32a + b + c = \frac{3}{2}, then the value of aa is
View answer and explanation
Correct option: D
12−12\frac{1}{2} - \frac{1}{{\sqrt 2 }}
Option D (12−12\frac{1}{2} - \frac{1}{{\sqrt 2 }}) is correct because solving the given conditions for AP and GP leads to this value for a.
743 Medium
743 of 998
Analyzing +4 | -1
{an}  and  {bn}\left\{ {{a_n}} \right\}\;and\;\left\{ {{b_n}} \right\} be two sequences given by an=(x)12n+(y)12n{a_n} = {\left( x \right)^{\frac{1}{{{2^n}}}}} + {\left( y \right)^{\frac{1}{{{2^n}}}}} and bn=(x)12n−(y)12n{b_n} = {\left( x \right)^{\frac{1}{{{2^n}}}}} - {\left( y \right)^{\frac{1}{{{2^n}}}}} for all n∈N,n \in N, then a1a2a3…an{a_1}{a_2}{a_3} \ldots {a_n} is equal to
View answer and explanation
Correct option: C
x−ybn\frac{{x - y}}{{{b_n}}}
Option C (x−ybn\frac{{x - y}}{{{b_n}}}) is correct because the product telescopes, simplifying to x−ybn\frac{x-y}{b_n} using the difference of squares repeatedly.
744 Easy
744 of 998
Application +4 | -1
The nnth term of the sequence 212,1713,119,2023…2 \frac{1}{2}, 1 \frac{7}{13}, 1 \frac{1}{9}, \frac{20}{23} \ldots is
View answer and explanation
Correct option: C
205n+3\frac{20}{5 n+3}
Option C is correct because the nth term of the sequence can be found by taking the reciprocal of the terms, recognizing an A.P., and then finding the reciprocal of the nth term of that A.P.
745 Easy
745 of 998
Application +4 | -1
If the sum of the first 2 n terms of the A.P. 2,5,8, is equal to the sum of the first nn terms of the A.P. 57, 59, 61,...., then nn equals
View answer and explanation
Correct option: C
11
Option C (11) is correct because it is the value of n that equates the sum of the first 2n terms of the first A.P. to the sum of the first n terms of the second A.P.
746 Easy
746 of 998
Application +4 | -1
Let a1,a2,a3,…,an,…….a_1, a_2, a_3, \ldots, a_n, \ldots \ldots .. be a G.P such that a4a6=14\frac{a_4}{a_6}=\frac{1}{4} and a2+a5=216a_2+a_5=216. Then, a1=a_1=
View answer and explanation
Correct option: A
12 or, 1087\frac{108}{7}
Option A (12 or, 108/7) is correct because solving the equations for the given G.P. yields two possible values for the first term.
747 Easy
747 of 998
Application +4 | -1
The value of 1−log⁡2+(log⁡2)22!−(log⁡2)33!+…1 - \log 2 + \frac{{{{\left( {\log 2} \right)}^2}}}{{2!}} - \frac{{{{\left( {\log 2} \right)}^3}}}{{3!}} + \ldots is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the given series is the expansion of e−log2e^{-log2} which simplifies to 12\frac{1}{2}.
748 Medium
748 of 998
Application +4 | -1
If x=1+y+y2+…x=1+y+y^2+\ldots, then yy is
View answer and explanation
Correct option: C
x−1x\frac{x-1}{x}
Option C (x−1x\frac{x-1}{x}) is correct because the given equation represents an infinite geometric series which can be solved for y.
749 Easy
749 of 998
Application +4 | -1
The geometric mean of the numbers 3,32,33,…,3n3,3^2, 3^3, \ldots, 3^n is
View answer and explanation
Correct option: D
3(n+1)/23^{(n+1) / 2}
Option D (3(n+1)/23^{(n+1) / 2}) is correct because it is the result of calculating the geometric mean of a geometric progression.
750 Easy
750 of 998
Application +4 | -1
The number of terms of the A.P. 1,4,7,…1, 4, 7, \ldots that must be taken to obtain a sum of 715 is
View answer and explanation
Correct option: C
22
Option C (22) is correct because by using the arithmetic progression sum formula, we can find the number of terms needed to obtain a sum of 715.
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