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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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691 Easy
691 of 998
Application +4 | -1
If log⁡1227=a\log _{12} 27=a, then log⁡616=\log _6 16=
View answer and explanation
Correct option: B
4(3−a3+a)4\left(\frac{3-a}{3+a}\right)
Option B (4(3−a3+a)4\left(\frac{{3-a}}{{3+a}}\right)) is correct because it results from expressing log⁡616\log_6 16 in terms of log⁡1227=a\log_{12} 27 = a using change of base formulas.
692 Medium
692 of 998
Analyzing +4 | -1
Let the sequence, a1,a2,a3,…,  a2n,{a_1},{a_2},{a_3}, \ldots ,\;{a_{2n}}, form an AP, then a12−a22+a32−...+a2n−12−a2n2a_1^2 - a_2^2 + a_3^2 - ... + a_{2n - 1}^2 - a_{2n}^2 is equal to
View answer and explanation
Correct option: A
n2n−1(a12−a2n2)\frac{n}{{2n - 1}}\left( {a_1^2 - a_{2n}^2} \right)
Option A is correct because the given expression can be simplified using the properties of arithmetic progressions and factorization.
693 Easy
693 of 998
Application +4 | -1
Let aa, bb and cc be in GP with common ratio rr, where a≠0a \ne 0 and 0<r≤120 < r \le \frac{1}{2}. If 3a, 7b and 15c are the first three terms of an AP, then the 4th term of this AP is
View answer and explanation
Correct option: C
a
Option C (a) is correct because after using the properties of GP and AP, and solving for the common ratio, the 4th term of the AP is found to be a.
694 Easy
694 of 998
Application +4 | -1
If one geometric mean GG and two arithmetic means A1A_1 and A2A_2 are inserted between two given quantities, then (2A1−A2)(2A2−A1)=\left(2 A_1-A_2\right)\left(2 A_2-A_1\right)=
View answer and explanation
Correct option: C
G2G^2
Option C (G2G^2) is correct because, by expressing A1 and A2 in terms of a and b, the expression simplifies to ab, which is equal to G squared.
695 Medium
695 of 998
Application +4 | -1
The number of numbers lying between 100 and 500 that are divisible by 7 but not by 21 , is
View answer and explanation
Correct option: C
38
Option C (38) is correct because it is the number of integers between 100 and 500 divisible by 7 but not by 21, found by subtracting the number divisible by 21 from the number divisible by 7.
696 Easy
696 of 998
Application +4 | -1
If xn>xn−1>…>x2>x1>1,x_n>x_{n-1}>\ldots>x_2>x_1>1, the value of log⁡x1 log⁡x2 log⁡x3 ⋅⋅⋅log⁡xn xnxn−1.....x1{\log _{{x_1}}}\,{\log _{{x_2}}}\,{\log _{{x_3}}}\, \cdot \cdot \cdot {\log _{{x_n}}}\,{x_n}^{{x_{n - 1}}^{{.^{{.^{{.^{{.^{.{x_1}}}}}}}}}}} is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the expression simplifies to 1 due to the properties of logarithms and exponents.
697 Medium
697 of 998
Application +4 | -1
If 12+22+32+…+20032=(2003)(4007)(334)1^2+2^2+3^2+\ldots+2003^2=(2003)(4007)(334) and (1)(2003)+(2)(2002)+(3)(2001)(1)(2003)+(2)(2002)+(3)(2001)
+…+(2003)(1)=(2003)(334)(x)+\ldots+(2003)(1)=(2003)(334)(x)
then xx equals
View answer and explanation
Correct option: A
2005
Option A (2005) is correct because by adding the two given series and simplifying, we can solve for x and find that x = 2005.
698 Easy
698 of 998
Application +4 | -1
The solution of the equation log⁡7log⁡5(x+5+x)=0\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0 is
View answer and explanation
Correct option: C
4
Option C (4) is correct because substituting x=4 into the equation satisfies the logarithmic conditions and simplifies to a valid solution.
699 Easy
699 of 998
Application +4 | -1
If a1,a2,a3(a1>0)a_1, a_2, a_3\left(a_1>0\right) are three successive terms of a G.P. with common ratio rr, the value of for which a3>4a2−3a1a_3>4 a_2-3 a_1 holds is given by
View answer and explanation
Correct option: C
r>3r>3 or r<1r<1
Option C (r>3 or r<1) is correct because it represents the solution set of the inequality derived from the given condition.
700 Easy
700 of 998
Application +4 | -1
If x1,x2,….,xnx_{1}, x_{2}, \ldots ., x_{n} and 1h1,1h2,….1hn\frac{1}{h_{1}}, \frac{1}{h^{2}}, \ldots . \frac{1}{h_{n}} are two A.P's such that x3=h2=8x_{3}=h_{2}=8 and x8=h7=20x_{8}=h_{7}=20, then x5.h10x_{5} . h_{10} equals.
View answer and explanation
Correct option: A
2560
Option A (2560) is correct because it is the product of x_5 and h_10, calculated using the properties of arithmetic progressions.
701 Medium
701 of 998
Application +4 | -1
A polygon has 25 sides, the lengths of which starting from the smallest side are in AP. If the perimeter of the polygon is 2100 cm2100 \mathrm{~cm} and the length of the largest side 20 times that of the smallest, then the length of the smallest side and the common difference of an AP are
View answer and explanation
Correct option: A
8 cm,613 cm8 \mathrm{~cm}, 6 \frac{1}{3} \mathrm{~cm}
Option A (8 cm, 6 1/3 cm) is correct because these are the length of the smallest side and the common difference of the AP, calculated using the given perimeter and largest side conditions.
702 Easy
702 of 998
Application +4 | -1
The value of 49A+5B49^A+5^B, where A=1−log⁡72A=1-\log _7 2 and B=−log⁡54B=-\log _5 4 is
View answer and explanation
Correct option: C
12.512.5
Option C (12.5) is correct because it results from correctly evaluating the expression 49A+5B49^A + 5^B using logarithm properties.
703 Easy
703 of 998
Application +4 | -1
After inserting nn A.Ms between 2 and 38, the sum of the resulting progression is 200. The value of nn is
View answer and explanation
Correct option: B
8
Option B (8) is correct because the value of n is calculated using the formula for the sum of an arithmetic progression.
704 Medium
704 of 998
Understanding +4 | -1
ex−1−12(x−1)2+13(x−1)3−14(x−1)4+…{e^{x - 1 - \frac{1}{2}{{\left( {x - 1} \right)}^2} + \frac{1}{3}{{\left( {x - 1} \right)}^3} - \frac{1}{4}{{\left( {x - 1} \right)}^4} + \ldots }} is equal to
View answer and explanation
Correct option: C
xx
Option C is correct because the exponent simplifies to log(x) using the Taylor series expansion of log(1+x), and e^(log(x)) equals x.
705 Medium
705 of 998
Application +4 | -1
Let the sum of the first n\mathrm{n} terms of a non-constant A.P., a1,a2,a3,…\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3, \ldots. be 50n+n(n−7)2 A50 \mathrm{n}+\frac{\mathrm{n}(\mathrm{n}-7)}{2} \mathrm{~A}, where A\mathrm{A} is a constant. If d\mathrm{d} is the common difference of this A.P., then the ordered pair (d,a50)\left(\mathrm{d}, \mathrm{a}_{50}\right) is equal to
View answer and explanation
Correct option: A
(A,50+46 A)(\mathrm{A}, 50+46 \mathrm{~A})
Option A ((A, 50+46A)) is correct because the common difference and the 50th term of the A.P. are calculated using the given sum of the first n terms.
706 Medium
706 of 998
Application +4 | -1
The value of ∑r=1n1a+rx+a+(r−1)x is \sum\limits_{{r=1}}^n \frac{1}{\sqrt{a+r x}+\sqrt{a+(r-1) x}} \text { is }
View answer and explanation
Correct option: A
na+a+nx\frac{n}{\sqrt{a}+\sqrt{a+n x}}
Option A (na+a+nx\frac{n}{\sqrt{a}+\sqrt{a+n x}}) is correct because, after rationalizing the denominator and summing the terms, the expression simplifies to this result.
707 Easy
707 of 998
Application +4 | -1
A man saves ₹ 200 in each of the first three months of his service. In each of the subsequent months his saving increases by ₹ 40 more than the saving in the immediately previous month. His total savings from the start of service will be ₹11040₹ 11040 after
View answer and explanation
Correct option: D
21 months
Option D (21 months) is correct because by setting up an equation representing the total savings as an arithmetic progression and solving the resulting quadratic, we find the number of months to be 21.
708 Medium
708 of 998
Application +4 | -1
If the sum of nn terms of an APA P is 3n2+5n3 n^2+5 n, then the common difference is
View answer and explanation
Correct option: D
6
Option D (6) is correct because the common difference of an AP with sum Sn=3n2+5nS_n = 3n^2 + 5n is 2A=2(3)=62A = 2(3) = 6.
709 Medium
709 of 998
Analyzing +4 | -1
If an=1+12+13+14+15+…+12n−1,{a_n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots + \frac{1}{{{2^n} - 1}}, then
View answer and explanation
Correct option: A
a100<100{a_{100}} < 100
Option A (a100<100{a_{100}} < 100) is correct because the sum an{a_n} is always less than nn, as each term in the series is overestimated by replacing fractions with larger values.
710 Medium
710 of 998
Analyzing +4 | -1
If a,b,ca,b,c are in G.P. and a−b,c−aa - b,c - a and b−cb - c are in H.P., then a+4b+ca + 4b + c is equal to
View answer and explanation
Correct option: B
0
Option B (0) is correct because the given conditions lead to the equation a+4b+c=0a + 4b + c = 0.
711 Medium
711 of 998
Application +4 | -1
Suppose a,b,ca, b, c are distinct positive real numbers such that a,2b,3ca, 2 b, 3 c are in A.P. and a,b,ca, b, c are in G.P. The common ratio of G.P. is
View answer and explanation
Correct option: C
1/31 / 3
Option C (1/3) is correct because by using the conditions for arithmetic and geometric progressions, we can derive a relationship between a and c, leading to the common ratio of 1/3.
712 Medium
712 of 998
Application +4 | -1
If one geometric mean GG and two arithmetic means pp and qq are inserted between two numbers, then G2G^2 is equal to
View answer and explanation
Correct option: B
(2p−q)(2q−p)(2 p-q)(2 q-p)
Option B ((2p-q)(2q-p)) is correct because it represents the square of the geometric mean in terms of the arithmetic means.
713 Easy
713 of 998
Application +4 | -1
The sum of 7 terms of a GP 3,6,12,…3,6,12, \ldots is
View answer and explanation
Correct option: C
381
Option C (381) is correct because the sum of the first 7 terms of the GP 3,6,12,...3, 6, 12, ... is 381.
714 Easy
714 of 998
Application +4 | -1
Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220. If the second term in it is 12 , then its 4th 4^{\text {th }} term is:
View answer and explanation
Correct option: C
20
Option C (20) is correct because it is the 4th term of the A.P. that satisfies the given conditions where the sum of the first nine terms is between 200 and 220, and the second term is 12.
715 Easy
715 of 998
Application +4 | -1
The value of 21/4⋅41/8⋅81/16⋅161/32……2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \cdot 16^{1 / 32} \ldots \ldots, is
View answer and explanation
Correct option: B
22
Option B (2) is correct because the expression simplifies to 212^1 after evaluating the infinite series in the exponent.
716 Medium
716 of 998
Application +4 | -1
The four numbers are in AP, such that their sum is 50 and greatest of them is 4 times the least, are
View answer and explanation
Correct option: C
5,10,15,205,10,15,20
Option C (5,10,15,205,10,15,20) is correct because these four numbers are in AP, their sum is 50, and the greatest is 4 times the least.
717 Easy
717 of 998
Application +4 | -1
If 1+sin⁡x+sin⁡2x+...upto  ∞=4+23,0<x<π  and  x≠π21 + \sin x + {\sin ^2}x + ...\mathrm{upto}\;\infty = 4 + 2\sqrt 3 ,0 < x < \pi \;\mathrm{and}\;x \ne \frac{\pi }{2}, then xx is equal to
View answer and explanation
Correct option: C
π3,2π3\frac{\pi }{3},\frac{{2\pi }}{3}
Option C (π3,2π3\frac{\pi }{3},\frac{{2\pi }}{3}) is correct because solving the infinite geometric series and then finding the sine inverse gives these two solutions within the specified range.
718 Easy
718 of 998
Analyzing +4 | -1
Let aa and bb be roots of x2−3  x+p=0{x^2} - 3\;x + p = 0 and let cc and dd be the roots of x2−12  x+q=0,{x^2} - 12\;x + q = 0, where a,b,c,da,b,c,d form an increasing GP Then the ratio of q+p:q−pq + p:q - p is equal to
View answer and explanation
Correct option: C
17  :1517\;:15
Option C (17:1517:15) is correct because the ratio is derived from the relationships between the roots of the quadratic equations and the geometric progression.
719 Easy
719 of 998
Understanding +4 | -1
Statement I: If 'aa' is the first term and 'dd' is the common difference of an A.P., then its nth n^{\text {th }} term is given by an=a−(n−1)da_n=a-(n-1) d

Statement II: The sum SnS_n of nn terms of an A.P. with first term 'aa' and common difference 'd{d}' is given by Sn=n2{2a+(n−1)d}.{S}_{{n}}=\frac{{n}}{2}\{2 {a}+({n}-1) {d}\}.

Choose the correct option.
View answer and explanation
Correct option: B
Only II is true
Option B (Only II is true) is correct because statement I has an incorrect sign, while statement II correctly states the sum of n terms of an A.P.
720 Easy
720 of 998
Application +4 | -1
If a,  b,  ca,\;b,\;c are in AP, then 3a,  3b,  3c{3^a},\;{3^b},\;{3^c} shall be in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because if a, b, c are in AP, then 3^a, 3^b, 3^c are in GP.
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