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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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901 Easy
901 of 998
Application +4 | -1
If 1,log⁡9(31−x+2)1, \log _9\left(3^{1-x}+2\right) and log⁡3(4.3x−1)\log _3\left(4.3^x-1\right) are in APA P, then xx is equal to
View answer and explanation
Correct option: C
1−log⁡341-\log _3 4
Option C (1−log⁡341-\log _3 4) is correct because by using the properties of logarithms and solving the resulting quadratic equation, we find x=1−log⁡34x = 1 - \log_3 4.
902 Easy
902 of 998
Application +4 | -1
Let a1,a2,a3,…a_1, a_2, a_3, \ldots be terms are in AP\mathrm{AP}, if a1+a2+…+apa1+a2+…+aq=p2q2,p≠q\frac{a_1+a_2+\ldots+a_p}{a_1+a_2+\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 (1141\frac{11}{41}) is correct because by using the formula for the sum of an AP and substituting p=11 and q=41, we can find the ratio of a6/a21.
903 Easy
903 of 998
Analyzing +4 | -1
If 210+29⋅31+28⋅32+…+2×39+310=S−2112^{10}+2^{9} \cdot 3^{1}+2^{8} \cdot 3^{2}+\ldots+2 \times 3^{9}+3^{10}=S-2^{11} then S\mathrm{S} is equal to:
View answer and explanation
Correct option: B
3113^{11}
Option B (3113^{11}) is correct because the given series is a G.P., and by applying the G.P. sum formula and rearranging, we find S = 3113^{11}.
904 Easy
904 of 998
Application +4 | -1
If 1−13+15−17+19−111+⋯=π41-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots=\frac{\pi}{4}, then value of 11⋅3+15⋅7+19⋅11+⋯\frac{1}{1 \cdot 3}+\frac{1}{5 \cdot 7}+\frac{1}{9 \cdot 11}+\cdots is
View answer and explanation
Correct option: A
π8\frac{\pi}{8}
Option A (π8\frac{\pi}{8}) is correct because it is derived by recognizing the given series as a manipulation of the Leibniz formula for pi.
905 Easy
905 of 998
Application +4 | -1
If x=∑n=0∞an,y=∑n=0∞bn,z=∑n=0∞cnx=\sum_{n=0}^{\infty} a^n, y=\sum_{n=0}^{\infty} b^n, z=\sum_{n=0}^{\infty} c^n where a,b,ca, b, c are in A.PA . P. and ∣a∣<1,∣b∣<1,∣c∣<1,abc≠0|a|<1,|b|<1,|c|<1, a b c \neq 0, then
View answer and explanation
Correct option: C
1x,1y,1z\frac{1}{x}, \frac{1}{y}, \frac{1}{z} are in A.P.
Option C (1x,1y,1z\frac{1}{x}, \frac{1}{y}, \frac{1}{z} are in A.P.) is correct because it follows from the given conditions that a, b, c are in A.P., and a, b, c can be expressed in terms of x, y, z.
906 Easy
906 of 998
Application +4 | -1
The sum of all 2 digit odd numbers is
View answer and explanation
Correct option: A
2475
Option B (2475) is correct because it is the sum of all two-digit odd numbers, calculated using the arithmetic series formula.
907 Easy
907 of 998
Application +4 | -1
Suppose a,b,ca, b, c are positive and form an increasing G.P. If a+b+c=xba+b+c=x b, then
View answer and explanation
Correct option: A
x>3x>3
Option A (x>3) is correct because, given the conditions of the increasing G.P., x must be greater than 3.
908 Easy
908 of 998
Application +4 | -1
The value of (0.16)log⁡2.5(13+132+133+  …to  ∞){\left( {0.16} \right)^{{{\log }_{2.5}}\left( {\frac{1}{3} + \frac{1}{{{3^2}}} + \frac{1}{{{3^3}}} + \; \ldots \mathrm{to}\;\infty } \right)}} is
View answer and explanation
Correct option: D
4
Option D (4) is correct because after simplifying the logarithmic expression and applying the properties of exponents, the result is 4.
909 Easy
909 of 998
Application +4 | -1
The sum of the first and third term of an arithmetic series is 12 and the product of first and second term is 24, then first term is
View answer and explanation
Correct option: C
4
Option C (4) is correct because by setting up equations based on the given conditions for the arithmetic series, we can solve for the first term.
910 Easy
910 of 998
Application +4 | -1
Sum of the first 20terms20 \mathrm{terms} of the AP 2,314,412,534,…2,3 \frac{1}{4}, 4 \frac{1}{2}, 5 \frac{3}{4}, \ldots is
View answer and explanation
Correct option: B
27712277 \frac{1}{2}
Option B (27712277 \frac{1}{2}) is correct because it is the sum of the first 20 terms of the given arithmetic progression, calculated using the formula for the sum of an AP.
911 Easy
911 of 998
Application +4 | -1
If mm times the mmth term of an A.P. with non-zero common difference equals nn times the nnth term of the A.P., where m≠nm \neq n, then (m+n)(m+n)th term of this A.P. is
View answer and explanation
Correct option: B
Zero
The correct answer is Zero (Option B) because if m times the mth term equals n times the nth term, then the (m+n)th term of the A.P. is zero.
912 Easy
912 of 998
Remembering +4 | -1
∑m=1nm2\sum\limits_{{m=1}}^n m^2 is equal to
View answer and explanation
Correct option: C
n(n+1)(2n+1)6\frac{n(n+1)(2 n+1)}{6}
Option C (n(n+1)(2n+1)6\frac{n(n+1)(2 n+1)}{6}) is correct because it is the standard formula for the sum of the squares of the first n natural numbers.
913 Easy
913 of 998
Application +4 | -1
If a,b,c,d,ea, b, c, d, e are in AP then the value of a+b+4c−4d+ea+b+4 c-4 d+e in terms of aa, if possible is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expression a+b+4c−4d+ea+b+4c-4d+e cannot be expressed solely in terms of 'a' as the AP relationship introduces dependencies on the common difference.
914 Easy
914 of 998
Application +4 | -1
Sum to nn terms of the series 1(1+x)(1+2x)+1(1+2x)(1+3x)+1(1+3x)(1+4x)+…\frac{1}{(1+x)(1+2 x)}+\frac{1}{(1+2 x)(1+3 x)}+\frac{1}{(1+3 x)(1+4 x)}+\ldots is
View answer and explanation
Correct option: B
n(1+x)[1+(n+1)x]\frac{n}{(1+x)[1+(n+1) x]}
Option B (n(1+x)[1+(n+1)x]\frac{n}{(1+x)[1+(n+1) x]}) is correct because it is the sum to n terms of the given series, obtained by using partial fractions.
915 Easy
915 of 998
Application +4 | -1
The minimum value of 4x+41−x,x∈R4^x+4^{1-x}, x \in R, is
View answer and explanation
Correct option: B
4
Option B (4) is correct because the minimum value of the expression is found using AM >= GM inequality, resulting in a minimum value of 4.

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916 Easy
916 of 998
Application +4 | -1
If a,b,ca, b, c are in AP, then (a−c)2(b2−ac)=\frac{(a-c)^{2}}{\left(b^{2}-a c\right)}=
View answer and explanation
Correct option: D
4
Option D (4) is correct because if a, b, c are in AP, then 2b=a+c2b = a+c. Substituting this into the expression and simplifying gives 4.
917 Easy
917 of 998
Analyzing +4 | -1
The value of 23!+45!+67!+…\frac{2}{{3!}} + \frac{4}{{5!}} + \frac{6}{{7!}} + \ldots is
View answer and explanation
Correct option: B
e−1{e^{ - 1}}
Option B (e−1{e^{ - 1}}) is correct because the series can be rewritten as the expansion of e−1{e^{ - 1}}.
918 Easy
918 of 998
Application +4 | -1
Three successive terms of a G.P. will form the sides of a triangle if the common ratio rr satisfies the inequality
View answer and explanation
Correct option: B
5−12<r<5+12\frac{\sqrt{5}-1}{2} < r < \frac{\sqrt{5}+1}{2}
Option B (5−12<r<5+12\frac{\sqrt{5}-1}{2} < r < \frac{\sqrt{5}+1}{2}) is correct because it satisfies the triangle inequality theorem for the sides of a triangle formed by three successive terms of a G.P.
919 Easy
919 of 998
Analyzing +4 | -1
If 1(20−a)(40−a)+1(40−a)(60−a)+…+\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+ 1(180−a)(200−a)=1256\frac{1}{(180-a)(200-a)}=\frac{1}{256}, then the maximum value of aa is :
View answer and explanation
Correct option: C
212
Option C (212) is correct because solving the given equation leads to a quadratic equation in 'a', and the maximum value of 'a' is 212.
920 Easy
920 of 998
Analyzing +4 | -1
The following consecutive terms 11+x,11−x,11−x\frac{1}{{1 + \sqrt x }},\frac{1}{{1 - x}},\frac{1}{{1 - \sqrt x }} of a series are in
View answer and explanation
Correct option: C
A.P.
Option C (A.P.) is correct because the difference between consecutive terms is constant.
921 Easy
921 of 998
Analyzing +4 | -1
The sum of nn terms of the series
12⋅2213+22⋅3213+23+32⋅4213+23+33+….\frac{\frac{1}{2} \cdot \frac{2}{2}}{1^3}+\frac{\frac{2}{2} \cdot \frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2} \cdot \frac{4}{2}}{1^3+2^3+3^3}+\ldots .. Is
View answer and explanation
Correct option: B
nn+1\frac{\mathrm{n}}{\mathrm{n}+1}
Option B (n/(n+1)) is correct because the nth term can be simplified and the sum telescopes, resulting in n/(n+1).
922 Easy
922 of 998
Analyzing +4 | -1
The sum of infinite terms of the series 11.4.7+14.7.10+……⋅1(3n−2)(3n+1)(3n+4)\frac{1}{1.4 .7}+\frac{1}{4.7 .10}+\ldots \ldots \cdot \frac{1}{(3 n-2)(3 n+1)(3 n+4)} is
View answer and explanation
Correct option: B
124\frac{1}{24}
Option B (1/24) is correct because the sum of the infinite series can be found using the formula for telescoping series.
923 Easy
923 of 998
Application +4 | -1
If the sides of sides of a right angled triangle are in AP, then the sides are proportional to
View answer and explanation
Correct option: C
3:4:53\mathrm{ }:\mathrm{ }4\mathrm{ }:\mathrm{ }5
Option C (3:4:53\mathrm{ }:\mathrm{ }4\mathrm{ }:\mathrm{ }5) is correct because it represents the sides of a right-angled triangle in AP, satisfying the Pythagorean theorem.
924 Easy
924 of 998
Application +4 | -1
There exist a positive number λ\lambda, such that log⁡2x+log⁡4x+log⁡8x=log⁡λx\log _2 x+\log _4 x+\log _8 x=\log _\lambda x, for all positive real numbers xx.

If λ=ab\lambda=\sqrt[b]{a}, where a,b∈Na, b \in N, the smallest possible value of (a+b)(a+b) is equal to
View answer and explanation
Correct option: D
75
Option D (75) is correct because simplifying the logarithmic equation leads to λ=26/11\lambda = 2^{6/11}, and expressing this in the form λ=ab\lambda = \sqrt[b]{a} gives a=26=64a = 2^6 = 64 and b=11b = 11, so a+b=75a + b = 75.
925 Easy
925 of 998
Analyzing +4 | -1
The sum of the series (1+2)(1+2+22)+(1+2+22+23)+…\left( {1 + 2} \right)\left( {1 + 2 + {2^2}} \right) + \left( {1 + 2 + {2^2} + {2^3}} \right) + \ldots up to nn terms is
View answer and explanation
Correct option: A
2n+2−n−4{2^{n + 2}} - n - 4
Option A (2n+2−n−4{2^{n + 2}} - n - 4) is correct because the sum of the series can be simplified using the formula for the sum of a geometric progression.
926 Easy
926 of 998
Application +4 | -1
Suppose (m+n)(m + n)th term of a G.P. is pp and (m–n)(m – n)th term is q,q, then its nnth is
View answer and explanation
Correct option: C
p(qp)m/2np{\left( {\frac{q}{p}} \right)^{m/2n}}
Option C (p(qp)m/2np{\left( {\frac{q}{p}} \right)^{m/2n}}) is correct because by using the properties of geometric progressions, we can find the nth term.
927 Easy
927 of 998
Application +4 | -1
If the second term of a GP is 2 and the sum of its infinite terms is 8 , then its first term is
View answer and explanation
Correct option: D
4
Option D (4) is correct because by using the given information about the second term and the sum of the infinite terms, we can set up a system of equations and solve for the first term, which equals 4.
928 Easy
928 of 998
Application +4 | -1
Let a1,a2,….,a10a_1, a_2, \ldots ., a_{10} be a G.P. If a3a1=25\frac{a_3}{a_1}=25, then a9a5\frac{a_9}{a_5} equals
View answer and explanation
Correct option: C
545^4
Option C (545^4) is the correct answer because, in a geometric progression, the ratio of terms equidistant from each other is constant, and this ratio can be calculated from the given information.
929 Easy
929 of 998
Application +4 | -1
Sum of the series ∑r=12016(−1)r(a+rd)\sum\limits_{{r=1}}^{2016}(-1)^r(a+r d) is
View answer and explanation
Correct option: D
1008d1008 d
Option D (1008d1008 d) is correct because the sum simplifies to 1008d1008d after evaluating the series.
930 Easy
930 of 998
Analyzing +4 | -1
The sum of series 12!−13!+14!−……\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots. upto infinity is
View answer and explanation
Correct option: D
e−1\mathrm{e}^{-1}
Option D (e⁻¹) is correct because the given series is the expansion of e^x with x = -1, minus the first two terms (1 and -1) which cancel each other.
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