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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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841 Easy
841 of 998
Application +4 | -1
If ppth term of an arithmetic progression is qq and the qqth term is pp, then 10th term is
View answer and explanation
Correct option: D
p+q−10p + q - 10
Option D is correct because by solving the system of equations formed by the given conditions, we can find the first term and common difference, and then determine the 10th term.
842 Easy
842 of 998
Application +4 | -1
Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is
View answer and explanation
Correct option: B
2+32+\sqrt{3}
Option D (126(350−1)\frac{{1}}{{26}}(3^{50}-1)) is correct because it is the sum of the first 50 terms of the G.P. calculated using the given conditions.
843 Medium
843 of 998
Application +4 | -1
The coefficient of n−r{n^{ - r}} in the expansion of log⁡10(nn−1),{\log _{10}}\left( {\frac{n}{{n - 1}}} \right), is
View answer and explanation
Correct option: A
1rlog⁡e10\frac{1}{{r{{\log }_e}10}}
Option A is correct because the coefficient of n−r{n^{-r}} is found by expanding the logarithm and identifying the term associated with n−r{n^{-r}}.
844 Easy
844 of 998
Application +4 | -1
If tan⁡nθ=tan⁡mθ,\tan n\mathrm{\theta } = \tan m\mathrm{\theta }, then the different values of θ\mathrm{\theta } will be
View answer and explanation
Correct option: A
AP
Option A (AP) is correct because the different values of θ\theta form an arithmetic progression.
845 Medium
845 of 998
Application +4 | -1
If f(x)=cos⁡2x+sec⁡2xf\left( x \right) = {\cos ^2}x + {\sec ^2}x, then
View answer and explanation
Correct option: D
f(x)≥2f\left( x \right) \ge 2
Option D is correct because the minimum value of cos²x + sec²x is 2, which is achieved when cos²x = 1.
846 Easy
846 of 998
Application +4 | -1
The value of log⁡497+log255−log⁡42log⁡17.5\frac{{\log 49\sqrt 7 + \mathrm{log}25\sqrt 5 - \log 4\sqrt 2 }}{{\log 17.5}} is
View answer and explanation
Correct option: C
52\frac{5}{2}
Option C (52\frac{5}{2}) is correct because simplifying the logarithmic expression results in 52\frac{5}{2}.
847 Easy
847 of 998
Application +4 | -1
If log⁡6(x+3)−log⁡6x=2,{\log _6}\left( {x + 3} \right) - {\log _6}x = 2, then x=x =
View answer and explanation
Correct option: B
335\frac{3}{{35}}
Option B (335\frac{3}{{35}}) is correct because it is found by solving the logarithmic equation using properties of logarithms.
848 Easy
848 of 998
Analyzing +4 | -1
Let Sn=113+1+213+23+...+1+2+...+n13+23+...+n3,  n=1,2,3,…{S_n} = \frac{1}{{{1^3}}} + \frac{{1 + 2}}{{{1^3} + {2^3}}} + ... + \frac{{1 + 2 + ... + n}}{{{1^3} + {2^3} + ... + {n^3}}},\;n = 1,2,3, \ldots Then Sn{S_n} is not greater than
View answer and explanation
Correct option: C
2
Option C (2) is correct because the sum of the series Sn{S_n} is always less than or equal to 2, as shown by simplifying the general term and summing the series.
849 Easy
849 of 998
Analyzing +4 | -1
The sum of the series 121⋅2!+12+222⋅3!+12+22+323⋅4!+…+12+22+…+n2n⋅(n+1)!+…∞\frac{1^2}{1 \cdot 2 !}+\frac{1^2+2^2}{2 \cdot 3 !}+\frac{1^2+2^2+3^2}{3 \cdot 4 !}+\ldots+\frac{1^2+2^2+\ldots+n^2}{n \cdot(n+1) !}+\ldots \infty: Equals
View answer and explanation
Correct option: C
3e−16\frac{3 e-1}{6}
Option C ((3e-1)/6) is correct because the sum of the given infinite series can be simplified and expressed in terms of the exponential constant 'e'.
850 Medium
850 of 998
Application +4 | -1
If the nnth term of a series is (nx)+y\left(\frac{n}{x}\right)+y, then sum of rr terms will be
View answer and explanation
Correct option: A
{r(r+1)2x}+ry\left\{\frac{r(r+1)}{2 x}\right\}+r y
Option A (\left{\frac{{r(r+1)}}{{2x}}\right}+ry\left{\frac{{r(r+1)}}{{2x}}\right}+ry) is correct because it is the sum of an arithmetic progression with the given nth term.
851 Medium
851 of 998
Analyzing +4 | -1
(aa+x)12+(aa−x)12=\left(\frac{a}{a+x}\right)^{\frac{1}{2}}+\left(\frac{a}{a-x}\right)^{\frac{1}{2}}= ?
View answer and explanation
Correct option: A
2+3x24a2+…2+\frac{3 x^2}{4 a^2}+\ldots
Option A (2+3x24a2+…2+\frac{3 x^2}{4 a^2}+\ldots) is correct because expanding both terms using the binomial theorem and adding them together results in the given expression.
852 Easy
852 of 998
Analyzing +4 | -1
If S1,S2{S_1},{S_2} and S3{S_3} denote the sum of first n1,n2{n_1},{n_2} and n3{n_3} terms respectively of an A.P., then S1n1(n2−n3)+S2n2(n3−n1)+S3n3(n1−n2)=\frac{{{S_1}}}{{{n_1}}}\left( {{n_2} - {n_3}} \right) + \frac{{{S_2}}}{{{n_2}}}\left( {{n_3} - {n_1}} \right) + \frac{{{S_3}}}{{{n_3}}}\left( {{n_1} - {n_2}} \right) =
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given expression simplifies to zero when substituting the formulas for the sum of an arithmetic progression.
853 Medium
853 of 998
Application +4 | -1
The sum to n terms of the infinite series 1.32+2.52+3.72+...∞{1.3^2} + {2.5^2} + {3.7^2} + ...\infty is
View answer and explanation
Correct option: A
n6(n+1)(6n2+14n+7)\frac{n}{6}\left( {n + 1} \right)\left( {6{n^2} + 14n + 7} \right)
Option A is correct because the sum to n terms of the given infinite series is derived by finding the nth term and then summing the series using standard summation formulas.
854 Easy
854 of 998
Application +4 | -1
If 1p+q,1q+r\frac{1}{p+q}, \frac{1}{q+r} and 1r+p\frac{1}{r+p} are in AP\mathrm{AP}, then
View answer and explanation
Correct option: B
q2,p2q^2, p^2 and r2r^2 are in AP
Option B (q2,p2,r2q^2, p^2, r^2 are in AP) is correct because it is derived from the given AP condition on the reciprocals of (p+q), (q+r), and (r+p).
855 Easy
855 of 998
Understanding +4 | -1
The coefficient of x99x^{99} in the expansion of (x−1)(x−2)……(x−100)(x-1)(x-2) \ldots \ldots(x-100), is
View answer and explanation
Correct option: C
−5050-5050
Option C (-5050) is correct because the coefficient of x99x^{{99}} is the negative sum of the roots of the polynomial.

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856 Easy
856 of 998
Application +4 | -1
If a,b,ca, b, c and dd are four positive real numbers such that abcd=1a b c d=1, the minimum value of (1+a)(1+b)(1+c)(1+d)(1+a)(1+b)(1+c)(1+d) is
View answer and explanation
Correct option: C
16
Option B (3a1a53 a_1 a_5) is correct because the expression a1a2+a2a3+a3a4+a4a5a_1 a_2+a_2 a_3+a_3 a_4+a_4 a_5 simplifies to 4a1a54a_1 a_5 when a1,a2,a3,a4,a5a_1, a_2, a_3, a_4, a_5 are in Harmonic Progression.
857 Easy
857 of 998
Application +4 | -1
If aa and xx are positive integers such that x<ax < a and a−x,x,a+x\sqrt{a-x}, \sqrt{x}, \sqrt{a+x} are in A.P., then least possible value of aa is
View answer and explanation
Correct option: A
5
Option A (5) is correct because it is the smallest possible integer value for 'a' that satisfies the given conditions for the arithmetic progression.
858 Easy
858 of 998
Application +4 | -1
The sum of first nn terms of the series 12+2(22)+32+2(42)+52+2(62)+…1^2+2\left(2^2\right)+3^2+2\left(4^2\right)+5^2+2\left(6^2\right)+\ldots is n(n+1)2/2n(n+1)^2 / 2 when nn is even. When nn is odd, the sum is
View answer and explanation
Correct option: B
n2(n+1)/2n^2(n+1) / 2
Option C (13\frac{{1}}{{3}}) is correct. By setting up the equations based on the given information and solving for the common ratio, we find r = 1/3.
859 Easy
859 of 998
Application +4 | -1
If x32(log⁡2x−3)=18,{x^{\frac{3}{2}\left( {{{\log }_2}x - 3} \right)}} = \frac{1}{8}, then xx equals to
View answer and explanation
Correct option: A
2
Option A (2) is correct because it is the solution to the given exponential equation after applying logarithms and solving the resulting quadratic equation.
860 Easy
860 of 998
Application +4 | -1
If log⁡a3=log⁡b4=log⁡c5,\frac{{\log a}}{3} = \frac{{\log b}}{4} = \frac{{\log c}}{5}, then caca equals
View answer and explanation
Correct option: B
b2{b^2}
Option B (b2{b^2}) is correct because it follows directly from the properties of logarithms and exponentials.
861 Medium
861 of 998
Application +4 | -1
Three non-zero numbers a,ba, b and cc are in AP. If increasing aa by ∥\| or increasing cc by 2 , the numbers are in GP, then bb equals
View answer and explanation
Correct option: C
12
Option C (12) is correct because solving the equations derived from the AP and GP conditions leads to b = 12.
862 Medium
862 of 998
Application +4 | -1
A G.P. consists of an even number of terms. If terms sum of all the terms is 55 times the sum of the terms occupying odd places, the common ration will be equal to
View answer and explanation
Correct option: C
4
Option C (4) is correct because the common ratio is found by setting up an equation relating the sum of all terms to the sum of terms in odd places in the G.P.
863 Easy
863 of 998
Application +4 | -1
If 2x.32x=100,{2^x}{.3^{2x}} = 100, then xx belongs to
View answer and explanation
Correct option: C
(1,  2)\left( {1,\;2} \right)
Option C is correct because by simplifying the expression and taking the logarithm, we find that x lies between 1 and 2.
864 Medium
864 of 998
Application +4 | -1
If (a,b),(c,d),(e,f)(a, b),(c, d),(e, f) are the vertices of a triangle such that a,c,ea, c, e are in GP with common ratio rr and b,d,fb, d, f are in GP with common ratio ss, then the area of the triangle is
View answer and explanation
Correct option: C
ab2(r−1)(s−1)(s−r)\frac{a b}{2}(r-1)(s-1)(s-r)
Option C (ab2(r−1)(s−1)(s−r)\frac{{a b}}{{2}}(r-1)(s-1)(s-r)) is correct because the area of the triangle can be calculated using the determinant formula with the given vertices, and simplifying the determinant leads to the expression.
865 Easy
865 of 998
Application +4 | -1
If 32sin⁡ 2α−1,   14   and  34−2sin⁡ 2α{3^{2\sin \,2\alpha - 1}},\,\,\,14\,\,\,\mathrm{and }\,\,{3^{4 - 2\sin \,2\alpha }} are the first three terms of an APAP for some α,\alpha , then the sixth term of this APAP is
View answer and explanation
Correct option: D
66
The correct answer is D (66) because it is the sixth term of the arithmetic progression, calculated using the given condition that the first three terms form an AP.
866 Medium
866 of 998
Application +4 | -1
The sum of first 12 terms of the series 131+13+231+3+13+23+331+3+5+…\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\ldots to nn terms is equal to
View answer and explanation
Correct option: C
4092\frac{409}{2}
Option C (409/2) is correct because the sum of the first 12 terms of the series is calculated to be 409/2.
867 Easy
867 of 998
Application +4 | -1
The value of 21/4⋅41/8⋅81/16…∞2^{1 / 4} \cdot 4^{1 / 8} \cdot 8^{1 / 16} \ldots \infty is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the infinite product can be simplified to a power of 2, where the exponent is the sum of an infinite series.
868 Medium
868 of 998
Application +4 | -1
If y=2x2−1,y = 2{x^2} - 1, then 1x2+12x4+13x6+…∞\frac{1}{{{x^2}}} + \frac{1}{{2{x^4}}} + \frac{1}{{3{x^6}}} + \ldots \infty equals to
View answer and explanation
Correct option: A
log⁡e(y+1y−1){\log _e}\left( {\frac{{y + 1}}{{y - 1}}} \right)
Option A (log⁡e(y+1y−1)\log _e\left( {\frac{{y + 1}}{{y - 1}}} \right)) is correct because it is the result of simplifying the given series using logarithmic properties and the given substitution.
869 Medium
869 of 998
Analyzing +4 | -1
Let a1,a2,a3,…a_{1}, a_{2}, a_{3}, \ldots be an A.P. If a7=3a_{7}=3, the product a1a4a_{1} a_{4} is minimum and the sum of its first nn terms is zero, then n!−4an(n+2)n !-4 a_{n}(n+2) is equal to :
View answer and explanation
Correct option: B
24
The correct answer is 24 because after finding the values of a and d, and subsequently n, the expression n! - 4*a_n*(n+2) evaluates to 24.
870 Medium
870 of 998
Understanding +4 | -1
If 3,32,333,3^2, 3^3 are in GPG P, then log⁡3,log⁡9\log 3, \log 9 and log⁡27\log 27 are in
View answer and explanation
Correct option: A
APA P
Option A (APA P) is correct because the logarithms of terms in a geometric progression form an arithmetic progression.
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