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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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Showing 451-480 of 564 questions

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451 Medium
451 of 564
Application +4 | -1
If the sides of a right angled triangle are in GP, then the cosine of the greater acute angle is
View answer and explanation
Correct option: A
21+5\frac{2}{1+\sqrt{5}}
Option A (\frac{{2}}{{1+5\sqrt{{5}}}}}) is correct because it is the cosine of the greater acute angle when the sides of a right-angled triangle are in GP.
452 Medium
452 of 564
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}+\ldots \ldots \infty=\frac{\pi}{4}, then the value of 11.3+15.7+19.11+……\frac{1}{1.3}+\frac{1}{5.7}+\frac{1}{9.11}+\ldots \ldots is
View answer and explanation
Correct option: A
π8\frac{\pi}{8}
Option A (π8\frac{\pi}{8}) is correct because the given series can be manipulated to match the Leibniz formula for π/4\pi/4, and the desired sum is half of that.
453 Medium
453 of 564
Application +4 | -1
If in a GP of 3 nterms, S1S_1 denotes the sum of the first nn terms, S2S_2 denotes the sum of the second block of nn terms and S˙3\dot{S}_3 denotes the sum of the last nn terms, then S1,S2S_1, S_2 and S3S_3 are in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because the sums of consecutive blocks of n terms in a GP also form a GP.
454 Medium
454 of 564
Analyzing +4 | -1
If a,  b,  ca,\;b,\;c are three distinct positive real numbers which are in HP, then 3a+2b2a−b+3c+2b2c−b\frac{{3a + 2b}}{{2a - b}} + \frac{{3c + 2b}}{{2c - b}} is
View answer and explanation
Correct option: A
Greater than or equal to 10
Option A (Greater than or equal to 10) is correct because the expression simplifies to 10 plus a positive term, making it greater than 10.
455 Medium
455 of 564
Application +4 | -1
If 0.272727…,x0.272727 \ldots, x and 0.727272…0.727272 \ldots are in HP, then xx must be
View answer and explanation
Correct option: A
rational
Option A (rational) is correct because the Harmonic Mean of two rational numbers is always a rational number.
456 Medium
456 of 564
Application +4 | -1
If SS be the sum, PP be the product and RR be the sum of the reciprocals of nn terms of a GP, then P2{P^2} is equal to
View answer and explanation
Correct option: A
(SR)n{\left( {\frac{S}{R}} \right)^n}
Option A ((SR)n{( \frac{S}{R} )}^n) is correct because it is derived by expressing S, P, and R in terms of the first term and common ratio of the G.P. and then verifying the relationship.
457 Medium
457 of 564
Application +4 | -1
The sum of nn terms of the series 11+3+13+5+15+7+…\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\ldots is
View answer and explanation
Correct option: B
−1+2n+12\frac{-1+\sqrt{2 n+1}}{2}
Option B (−1+2n+12\frac{-1+\sqrt{2 n+1}}{2}) is correct because it is the sum of the given series after rationalizing and simplifying.
458 Medium
458 of 564
Application +4 | -1
The sum of the first nn terms of the series 12+34+78+1516+…\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots is
View answer and explanation
Correct option: C
n+2−n−1n+2^{-n}-1
Option C is correct because the sum of the series is found by rewriting each term as (1 - fraction) and then summing the resulting series.
459 Medium
459 of 564
Application +4 | -1
If the sum of nn terms of an AP is 3n2+5n3 n^2+5 n, then which of its terms is 164 ?
View answer and explanation
Correct option: B
27 th
Option B (27 th) is correct because it is the term number that has a value of 164 in the given arithmetic progression.
460 Medium
460 of 564
Analyzing +4 | -1
If A1,  A2,  A3{A_1},\,\,{A_2},\,\,{A_3} denote respectively the areas of an inscribed polygon of 2n2n sides, inscribed polygon of nn sides and circumscribed polygon of nn sides, then A2,  A1,  A3{A_2},\,\,{A_1},\,\,{A_3} are in
View answer and explanation
Correct option: B
G.P.
Option B (G.P.) is correct because the areas A2,A1,A3{A_2}, {A_1}, {A_3} form a geometric progression, as shown by the relationship A12=A2A3{A_1}^2 = {A_2}{A_3}.
461 Medium
461 of 564
Application +4 | -1
The sum of first nn terms of 312+512+22+712+22+32+…\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots is
View answer and explanation
Correct option: B
6nn+1\frac{6 n}{n+1}
Option B (6nn+1\frac{{6n}}{{n+1}}) is correct because it is the result of summing the series after expressing each term as a difference.
462 Medium
462 of 564
Analyzing +4 | -1
If an=−24n2−16n+5a_n=\frac{-2}{4 n^2-16 n+5}, then a1+a2+….+a25a_1+a_2+\ldots .+a_{25} is equal to
View answer and explanation
Correct option: D
50141\frac{50}{141}
Option D (50141\frac{{50}}{{141}}) is correct because the sum telescopes, leaving only the first and last terms, which simplify to the given fraction.
463 Medium
463 of 564
Application +4 | -1
The sum of the infinite series 11⋅4+14⋅7+17⋅10+⋯\frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\cdots is
View answer and explanation
Correct option: A
13\frac{1}{3}
Option A (13\frac{1}{3}) is correct because the sum of the infinite series is calculated by expressing the terms as a difference and taking the limit.
464 Medium
464 of 564
Application +4 | -1
The sum of nn terms of the series 1+(1+x)+(1+x+x2)+(1+x+x2+x3)+…1+(1+x)+\left(1+x+x^2\right)+\left(1+x+x^2+x^3\right)+\ldots. where x≠1x \neq 1 is
View answer and explanation
Correct option: C
n1−x−x(1−xn)(1−x)2\frac{n}{1-x}-\frac{x\left(1-x^n\right)}{(1-x)^2}
Option C (n1−x−x(1−xn)(1−x)2\frac{n}{1-x}-\frac{x(1-x^n)}{(1-x)^2}) is correct because it is derived by summing the geometric series within each term and then summing the resulting series.
465 Medium
465 of 564
Application +4 | -1
If the harmonic mean between aa and bb is HH, then the value of 1H−a+1H−b\frac{1}{H-a}+\frac{1}{H-b} is
View answer and explanation
Correct option: C
1a+1b\frac{1}{a}+\frac{1}{b}
Option C (1a+1b\frac{1}{a}+\frac{1}{b}) is correct because substituting the formula for H and simplifying the expression leads to the given result.

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466 Medium
466 of 564
Analyzing +4 | -1
If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratio of all such GPs is
View answer and explanation
Correct option: C
3
Option C (3) is correct because the sum of the common ratios of all such GPs is found by solving for the terms and then summing the possible common ratios.
467 Medium
467 of 564
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}}.
468 Medium
468 of 564
Application +4 | -1
If a1,a2,a3,…,an,…a_1, a_2, a_3, \ldots, a_n, \ldots are in A.P. such that a4−a7+a10=ma_4-a_7+a_{10}=m, then the sum of first 13 terms of this A.P.. is :
View answer and explanation
Correct option: C
13 m13 \mathrm{~m}
Option C (13m) is correct because using the properties of arithmetic progressions, we can express the sum of the first 13 terms in terms of the given expression a4−a7+a10=ma_4 - a_7 + a_{10} = m.
469 Medium
469 of 564
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.
470 Medium
470 of 564
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.
471 Medium
471 of 564
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.
472 Medium
472 of 564
Application +4 | -1
If the ratio of the sum of nn terms of two AP′sA P^{\prime} s is 2n:(n+1)2 n:(n+1), then ratio of their 8th 8^{\text {th }} terms is-
View answer and explanation
Correct option: A
15:815: 8
Option A (15: 8) is correct because it is the ratio of the 8th terms obtained by substituting the appropriate value of n in the given ratio of sums.
473 Medium
473 of 564
Analyzing +4 | -1
The sum of 11+12+14+21+22+24+⋯∞\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots \infty terms equals to:
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because the infinite sum telescopes after partial fraction decomposition, resulting in a value of 12\frac{1}{2}.
474 Medium
474 of 564
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.
475 Medium
475 of 564
Application +4 | -1
The number of terms of the series 26,21,16,11,…26,21,16,11, \ldots to be added so as to get the sum 74 is
View answer and explanation
Correct option: B
4
Option A (4) is correct because it is the number of terms to be added in the series 26, 21, 16, 11,... to get the sum 74, calculated by solving a quadratic equation.
476 Medium
476 of 564
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.
477 Medium
477 of 564
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.
478 Medium
478 of 564
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.
479 Medium
479 of 564
Application +4 | -1
If the sum of first nn terms of an A.P. be equal to the sum of its first mm terms, (m≠n)(m \neq n), then the sum of its first (m+n)(m+n) terms will be
View answer and explanation
Correct option: A
0
Option A (0) is correct because if the sum of the first n terms equals the sum of the first m terms, then the sum of the first (m+n) terms is 0.
480 Medium
480 of 564
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.
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