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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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Showing 361-390 of 998 questions

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361 Easy
361 of 998
Application +4 | -1
Let S⊂(0,π)S \subset(0, \pi) denote the set of values of xx satisfying the equation
81+∣cos⁡x∣+cos⁡2x+∣cos⁡3x∣+…to∞=43{8^{1 + |\cos x| + {{\cos }^2}x + \left| {{{\cos }^3}x} \right| + \ldots \mathrm{ to }\infty }} = {4^3}
Then S=S=
View answer and explanation
Correct option: D
{π/3,2π/3}\{\pi / 3,2 \pi / 3\}
Option D ({π/3,2π/3}\{\pi / 3,2 \pi / 3\}) is correct because these are the only values of x in the given interval that satisfy the equation after simplifying the infinite series.
362 Easy
362 of 998
Application +4 | -1
If the sum of first 11 terms of an A.P. a1,a2,a3,…a_1, a_2, a_3, \ldots is 0 , and the sum of the A.P. a1,a3,a5,…,a23a_1, a_3, a_5, \ldots, a_{23} is ka1,a1≠0k a_1, a_1 \neq 0. Then kk is equal to
View answer and explanation
Correct option: C
−725-\frac{72}{5}
Option C (-725\frac{72}{5}) is correct because it is derived from setting up equations based on the given sums and solving for kk.
363 Easy
363 of 998
Application +4 | -1
Let Tr{T_r}, be rrth term of an AP whose first term is aa and common difference is dd. If for some positive integers m,n,m≠n,  Tm=1nm,n,m \ne n,\;{T_m} = \frac{1}{n} and Tn=1m{T_n} = \frac{1}{m}, then a−da - d equals
View answer and explanation
Correct option: A
0
Option A (0) is correct because solving the given conditions for an A.P. leads to a - d = 0.
364 Easy
364 of 998
Application +4 | -1
If a,  b,  ca,\;b,\;c are digits, then the rational number represented by 0. cababab…cababab \ldots is
View answer and explanation
Correct option: D
99c+10a+b990\frac{{99c + 10a + b}}{{990}}
Option D is correct because it accurately converts the repeating decimal into a rational number by setting up equations and solving for the repeating part.
365 Medium
365 of 998
Application +4 | -1
If sum of nn terms of an AP is 2n+3n22n + 3{n^2} then rrth term is
View answer and explanation
Correct option: C
6r−16r - 1
Option C (6r−16r - 1) is correct because it is derived by subtracting the sum of (n-1) terms from the sum of n terms to find the nth term.
366 Easy
366 of 998
Analyzing +4 | -1
If ∣x∣<1\left| x \right| < 1 and y=x−x22+x33−x44+...y = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ..., then xx is equal to
View answer and explanation
Correct option: C
y+y22!+y33!+...y + \frac{{{y^2}}}{{2!}} + \frac{{{y^3}}}{{3!}} + ...
Option C (y+y22!+y33!+...y + \frac{{{y^2}}}{{2!}} + \frac{{{y^3}}}{{3!}} + ...) is correct because it represents the exponential function ey−1e^y - 1, which is equal to x when y=log⁡(1+x)y = \log(1+x).
367 Medium
367 of 998
Application +4 | -1
If sum of first 20 terms of an A.P. is equal
to sum of first 30 terms of the A.P. then sum of the first 50
terms of the A.P. is
View answer and explanation
Correct option: B
0
Option B (0) is correct because if the sum of the first 20 terms equals the sum of the first 30 terms, then the sum of the first 50 terms is 0.
368 Easy
368 of 998
Application +4 | -1
The sum to nn terms of the series 113+1+213+23+1+2+313+23+33+…\frac{1}{1^3}+\frac{1+2}{1^3+2^3}+\frac{1+2+3}{1^3+2^3+3^3}+\ldots is
View answer and explanation
Correct option: C
2nn+1\frac{2 n}{n+1}
Option C (2nn+1\frac{{2n}}{{n+1}}) is correct because simplifying the general term of the series and summing it results in the given expression.
369 Easy
369 of 998
Application +4 | -1
log⁡2a3=log⁡2b4=log⁡2c5λ\frac{{{{\log }_2}a}}{3} = \frac{{{{\log }_2}b}}{4} = \frac{{{{\log }_2}c}}{{5\mathrm{\lambda }}} andand a−3b−4c=1{a^{ - 3}}{b^{ - 4}}c = 1 then λ=\mathrm{\lambda } =
View answer and explanation
Correct option: C
5
Option C (5) is correct because it is the value of lambda that satisfies the given logarithmic equation and the condition a−3b−4c=1{a^{ - 3}}{b^{ - 4}}c = 1.
370 Medium
370 of 998
Application +4 | -1
The sum of the integers from 1 to 100 which are not divisible by 3 or 5 is
View answer and explanation
Correct option: D
2632
Option D (2632) is correct because the sum of integers from 1 to 100 not divisible by 3 or 5 is calculated by subtracting the sums of multiples of 3 and 5, and adding back the sum of multiples of 15.
371 Easy
371 of 998
Application +4 | -1
If the AM and GM of roots of a quadratic equations are 8 and 5 respectively, then the quadratic equation will be
View answer and explanation
Correct option: C
x2−16x+25=0{x^2} - 16x + 25 = 0
Option C (x² - 16x + 25 = 0) is correct because it is the quadratic equation formed using the given arithmetic mean and geometric mean of the roots.
372 Medium
372 of 998
Application +4 | -1
If e5x+exe3x\frac{{{e^{5x}} + {e^x}}}{{{e^{3x}}}} is expand in a series of ascending powers of xx and nn is an odd natural number, then the coefficient of xn,{x^n}, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the expansion of the given expression contains only even powers of x, making the coefficient of any odd power of x equal to zero.
373 Medium
373 of 998
Application +4 | -1
The value of 34log⁡49+271log⁡369+811log⁡53,  is{3^{\frac{4}{{{{\log }_4}9}}}} + {27^{\frac{1}{{{{\log }_{36}}9}}}} + {81^{\frac{1}{{{{\log }_5}3}}}},\;\mathrm{is}
View answer and explanation
Correct option: C
857
Option C, 857, is correct because it is the result of simplifying the expression using logarithm and exponent rules.
374 Medium
374 of 998
Application +4 | -1
Sum of the series 34+536+7144+……\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\ldots \ldots upto nn terms is equal to
View answer and explanation
Correct option: B
n(n+2)(n+1)2\frac{n(n+2)}{(n+1)^2}
Option B (n(n+2)(n+1)2\frac{n(n+2)}{(n+1)^2}) is correct because it represents the sum of the series up to n terms, derived by expressing each term as a difference.
375 Medium
375 of 998
Application +4 | -1
If x1,x2,x3x_1, x_2, x_3 as well as y1,y2,y3y_1, y_2, y_3 are in GP with same common ratio, then the points (x1,y1),(x2,y2)\left(x_1, y_1\right),\left(x_2, y_2\right) and (x3,y3)\left(x_3, y_3\right)
View answer and explanation
Correct option: A
lie on a straight line
The correct answer is A (lie on a straight line) because the area formed by the points is zero, indicating collinearity.
376 Easy
376 of 998
Application +4 | -1
If 21−x+21+x,f(x),3x+3−x2^{1-x}+2^{1+x}, f(x), 3^x+3^{-x} are in A.P., then the minimum value of f(x)f(x) is
View answer and explanation
Correct option: C
3
Option C is correct because the minimum value of f(x) is found by using AM ≥ GM inequality on the given arithmetic progression, resulting in a minimum value of 3.
377 Medium
377 of 998
Application +4 | -1
If the lengths of sides of a right triangle are in AP, then the sines of the acute angle are
View answer and explanation
Correct option: A
35,45\frac{3}{5}, \frac{4}{5}
Option A (35,45\frac{3}{5}, \frac{4}{5}) is correct because the sines of the acute angles are calculated based on the sides of the right triangle being in arithmetic progression.
378 Medium
378 of 998
Application +4 | -1
Let s1,s2,s3…,s10s_{1}, s_{2}, s_{3} \ldots, s_{10} respectively be the sum of 12 terms of 10 A.Ps whose first terms are 1,2,3,….,101,2,3, \ldots ., 10 and the common differences are 1,3,5,…,191,3,5, \ldots, 19 respectively. Then ∑i=110si\sum\limits_{{i=1}}^{10} s_{i} is equal to
View answer and explanation
Correct option: C
7260
Option C (7260) is correct because it is the result of summing the sums of the 10 arithmetic progressions, using the formula for the sum of an AP and simplifying the summation.
379 Easy
379 of 998
Application +4 | -1
If a geometric progression consisting of positive terms, each term equals the sum of the next two terms, then the common ratio of this progression equals
View answer and explanation
Correct option: D
12(5−1)\frac{1}{2}(\sqrt{5}-1)
Option D (12(5−1)\frac{1}{2}(\sqrt{5}-1)) is correct because it is the positive root of the quadratic equation derived from the given condition of the geometric progression.
380 Medium
380 of 998
Application +4 | -1
The length of a side of a square is aa metre. A second square is formed by joining the mid-points of these square. Then, a third square is formed by joining the mid-points of the second square and so on. Then, the sum of the area of the squares which carried upto infinity, is
View answer and explanation
Correct option: B
2a22 a^2
Option B (2a22a^2) is correct because the sum of the areas forms an infinite geometric series with a common ratio of 1/2.
381 Medium
381 of 998
Application +4 | -1
If SrS_r denotes the sum of the first rr terms of an APA P, then S3r−Sr−1S2r−S2r−1\frac{S_{3 r}-S_{r-1}}{S_{2 r}-S_{2 r-1}} is equal to
View answer and explanation
Correct option: B
2r+12 r+1
Option B (2r+12r+1) is correct because simplifying the expression leads to 2r+12r+1.
382 Easy
382 of 998
Application +4 | -1
The sum of all natural numbers ′n′'n' such that 100 < n<200100\, < \,n < 200 and HCF (91, n) >1(91,\,n)\, > 1 is
View answer and explanation
Correct option: D
3121
Option D (3121) is correct because it is the sum of all natural numbers between 100 and 200 that have a HCF greater than 1 with 91.
383 Easy
383 of 998
Application +4 | -1
If 32sin⁡2θ−1,14,34−2sin⁡2θ3^{2 \sin 2 \theta-1}, 14,3^{4-2 \sin 2 \theta} are first three terms of an A.P., then its fifth term is given by
View answer and explanation
Correct option: C
53
Option B (n2(n+1)/2n^2(n+1) / 2) is correct. When n is odd, the sum of the series can be found by subtracting the last term from the sum when n is even.
384 Hard
384 of 998
Analyzing +4 | -1
The sum of the series 1+12+222  !+12+22+323  !+12+22+32+424  !+…,1 + \frac{{{1^2} + {2^2}}}{{2\;!}} + \frac{{{1^2} + {2^2} + {3^2}}}{{3\;!}} + \frac{{{1^2} + {2^2} + {3^2} + {4^2}}}{{4\;!}} + \ldots , is
View answer and explanation
Correct option: B
176e\frac{{17}}{6}e
Option B (176e\frac{{17}}{6}e) is correct because it is found by expressing the general term of the series, and then using the power series expansion of exe^x.
385 Medium
385 of 998
Application +4 | -1
If (1.05)50=11.658(1.05)^{50}=11.658, then ∑n=149(1.05)n\sum_{n=1}^{49}(1.05)^n equals
View answer and explanation
Correct option: C
212.16
Option C (212.16) is correct because the sum of the geometric series can be calculated using the formula for the sum of a GP, and substituting the given value of (1.05)50.
386 Easy
386 of 998
Application +4 | -1
If the product of nn positive numbers is unity, then their sum is
View answer and explanation
Correct option: D
never less than nn
Option D (never less than n) is correct because, by the AM-GM inequality, the sum of n positive numbers whose product is unity is always greater than or equal to n.
387 Easy
387 of 998
Application +4 | -1
Let aa and bb be two given numbers. If AA denotes the single AM and SS denotes the sum of nn AM's between aa and bb, then S/AS / A depends on
View answer and explanation
Correct option: D
nn
Option D (nn) is correct because the ratio S/A simplifies to n, showing it only depends on the number of arithmetic means.
388 Easy
388 of 998
Application +4 | -1
If a1,a2,a3,a4,a5{a_1},{a_2},{a_3},{a_4},{a_5} and a6{a_6} are six arithmetic means between 3 and 31, then a6−a5{a_6} - {a_5} and a1+a6{a_1} + {a_6} are respectively equals to
View answer and explanation
Correct option: C
4 and 34
Option C (4 and 34) is correct because the common difference is 4, and a6 - a5 = d = 4, and a1 + a6 = 7 + 27 = 34.
389 Easy
389 of 998
Understanding +4 | -1
∑j=1n∑i=1ni\sum\limits_{{j=1}}^{n} \sum\limits_{{i=1}}^{n} i is equal to
View answer and explanation
Correct option: C
n2(n+1)2\frac{n^{2}(n+1)}{2}
Option C (n2(n+1)2\frac{{n^{{2}}(n+1)}}{{2}}) is correct because the inner sum is n(n+1)2\frac{{n(n+1)}}{{2}}, and the outer sum multiplies this by n.
390 Easy
390 of 998
Application +4 | -1
The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternativety positive and negative, then the first term is
View answer and explanation
Correct option: D
-12
Option D (-12) is correct because it satisfies the given conditions of the geometric progression with alternating signs.
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