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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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421 Easy
421 of 998
Application +4 | -1
It is given that 114+124+134+...+to  ∞=π490\frac{1}{{{1^4}}} + \frac{1}{{{2^4}}} + \frac{1}{{{3^4}}} + ... + \mathrm{to}\;\infty = \frac{{{\pi ^4}}}{{90}}. Then, 114+134+154+...∞\frac{1}{{{1^4}}} + \frac{1}{{{3^4}}} + \frac{1}{{{5^4}}} + ...\infty is equal to
View answer and explanation
Correct option: A
π496\frac{{{\pi ^4}}}{{96}}
Option A (π496\frac{{{\pi ^4}}}{{96}}) is correct because it is derived by subtracting the sum of even terms from the total sum of the series.
422 Easy
422 of 998
Application +4 | -1
Insert two harmonic means between 3 and 5 .
View answer and explanation
Correct option: B
4513,4511\frac{45}{13}, \frac{45}{11}
Option B (45/13, 45/11) is correct because these are the two harmonic means between 3 and 5, calculated using the formula for inserting harmonic means.
423 Easy
423 of 998
Application +4 | -1
The value of log⁡5(1+15)+log⁡5(1+16)+log⁡5(1+17)+…{\log _5}\left( {1 + \frac{1}{5}} \right) + {\log _5}\left( {1 + \frac{1}{6}} \right) + {\log _5}\left( {1 + \frac{1}{7}} \right) + \ldots

+log⁡5(1+1624)  is+ {\log _5}\left( {1 + \frac{1}{{624}}} \right)\;\mathrm{is}
View answer and explanation
Correct option: C
3
Option C (3) is correct because the logarithmic expression simplifies to log⁡5(125){\log _5}(125), which equals 3.
424 Easy
424 of 998
Application +4 | -1
If a,b,ca,b,c are in G.P., then log⁡ax,log⁡bx,log⁡cx{\log _a}x,{\log _b}x,{\log _c}x are in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the reciprocals of the logarithms form an arithmetic progression, indicating that the original logarithms are in harmonic progression.
425 Easy
425 of 998
Application +4 | -1
The coefficient of xn{x^n} in the series

1+a+bx1!+(a+bx)22!+(a+bx)33!+…,1 + \frac{{a + bx}}{{1!}} + \frac{{{{\left( {a + bx} \right)}^2}}}{{2!}} + \frac{{{{\left( {a + bx} \right)}^3}}}{{3!}} + \ldots , is
View answer and explanation
Correct option: C
eabnn  !{e^a}\frac{{{b^n}}}{{n\;!}}
Option C (eabnn  !{e^a}\frac{{{b^n}}}{{n\;!}}) is correct because the coefficient of xn{x^n} in the given series is found by expanding the exponential function.
426 Easy
426 of 998
Application +4 | -1
If Hn=1+12+13+…+1nH_n=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}, then value of 1+32+53+…+2n−1n1+\frac{3}{2}+\frac{5}{3}+\ldots+\frac{2 n-1}{n} is
View answer and explanation
Correct option: B
2n−Hn2 n-H_n
Option B is correct because the given expression can be simplified using the definition of H_n.
427 Easy
427 of 998
Application +4 | -1
The sum ∑k=120k12k\sum_{k=1}^{20} k \frac{1}{2^{k}} is equal to :
View answer and explanation
Correct option: C
2−112192-\frac{11}{2^{19}}
Option C (2-11219\frac{11}{2^{19}}) is correct because it is the result of summing the given series using the method of differences.
428 Easy
428 of 998
Application +4 | -1
The sum to nn terms of the series 17+10+110+13+113+16+⋯\frac{1}{\sqrt{7}+\sqrt{10}}+\frac{1}{\sqrt{10}+\sqrt{13}}+\frac{1}{\sqrt{13}+\sqrt{16}}+\cdots is
View answer and explanation
Correct option: A
13(7+3n−7)\frac{1}{3}(\sqrt{7+3 n}-\sqrt{7})
Option A (13(7+3n−7)\frac{{1}}{{3}}(\sqrt{{7+3n}}-\sqrt{{7}})) is correct because rationalizing the denominator of each term and summing the telescoping series leads to this result.
429 Easy
429 of 998
Application +4 | -1
The value of ∑r=1nlog⁡(arbr−1)\sum \limits_{r = 1}^n \log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right) is
View answer and explanation
Correct option: C
n2log⁡(an+1bn−1)\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)
Option C (n2log⁡(an+1bn−1)\frac{n}{2}\log \left( {\frac{{{a^{n + 1}}}}{{{b^{n - 1}}}}} \right)) is correct because by applying logarithm properties and simplifying the summation, the expression is equivalent to the given option.
430 Easy
430 of 998
Application +4 | -1
Let two numbers have arithmetic mean 9 and geometric mean 4. Then, these numbers are the roots of the quadratic equation
View answer and explanation
Correct option: B
x2−18x+16=0x^2-18 x+16=0
Option B (x2−18x+16=0x^2-18 x+16=0) is correct because, given the arithmetic and geometric means, the quadratic equation is formed using the formula x2−2Ax+G2=0x^2 - 2Ax + G^2 = 0.
431 Medium
431 of 998
Application +4 | -1
If a,ba, b and cc are in HP, then the straight line xa+yb+1c=0\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0 always passes through a fixed point and that point is
View answer and explanation
Correct option: C
(1,−2)(1,-2)
Option C ((1,-2)) is correct because if a, b, and c are in HP, then 1/a, 1/b, and 1/c are in AP, leading to the condition that the line passes through the fixed point (1, -2).
432 Easy
432 of 998
Application +4 | -1
Let f:R→Rf: R \rightarrow R be a function which satisfies f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) for all x,y∈Rx, y \in R. If f(1)=2f(1)=2 and g(n)=∑k=1n−1f(k),n∈Ng(n)=\sum\limits_{k=1}^{n-1} f(k), n \in N, then the value of nn for which g(n)=20g(n)=20, is
View answer and explanation
Correct option: C
5
Option C (5) is correct because solving the summation and substituting into the given equation yields n=5.
433 Medium
433 of 998
Analyzing +4 | -1
The consecutive numbers of a three digit number form a GP. If we subtract 792 from this number, then we get a number consisting of the same digits written in the reverse order and if we increase the second digit of the required number by 2 , then resulting number forms an AP. The number is
View answer and explanation
Correct option: C
951
Option C (951) is correct because it satisfies the conditions of being a three-digit number in GP, reversing digits after subtracting 792, and forming an AP when the middle digit is increased by 2.
434 Easy
434 of 998
Application +4 | -1
The sum of 10 terms of the series 312×22+ 522×32+ 732×42+........\frac{3}{{{1^2} \times {2^2}}} + \,\frac{5}{{{2^2} \times {3^2}}} + \,\frac{7}{{{3^2} \times {4^2}}} + ........ is
View answer and explanation
Correct option: B
120/121
Option B (120/121) is correct because it is the sum of the 10 terms of the given series, calculated by expressing each term as a difference of two terms and then summing them.
435 Medium
435 of 998
Application +4 | -1
If f(x)f(x) is a function satisfying f(x+y)=f(x)f(y),∀x,y∈Nf(x+y)=f(x) f(y), \forall x, y \in N such that f(1)=3f(1)=3 and ∑x=1nf(x)=120\sum_{x=1}^n f(x)=120. Then, the value of nn is
View answer and explanation
Correct option: A
4
Option A (4) is correct because by using the given functional equation and the sum of the function values, we can determine that f(x) = 3^x, and solving the sum equation gives n = 4.
436 Medium
436 of 998
Analyzing +4 | -1
If Sn=∑r=0n1nCrS_n=\sum_{r=0}^n \frac{1}{{ }^n C_r} and tn=∑r=0vrrCrt_n=\sum_{r=0}^v \frac{r}{{ }^r C_r}, then tnSs\frac{t_n}{S_s} is equal to:
View answer and explanation
Correct option: D
12n\frac{1}{2} n
Option D (12n\frac{1}{2} n) is correct because by using the property nCr=nCn−r{ }^n C_r = { }^n C_{n-r} and manipulating the summation, we can show that tnSn=n2\frac{t_n}{S_n} = \frac{n}{2}.
437 Easy
437 of 998
Application +4 | -1
If nn is odd, the sum of nn terms of the series 12−22+32−42+52−62+……1^2-2^2+3^2-4^2+5^2-6^2+\ldots \ldots, is
View answer and explanation
Correct option: A
n(n+1)2\frac{n(n+1)}{2}
Option A (n(n+1)2\frac{n(n+1)}{2}) is correct because the sum of the series can be simplified to this expression when n is odd.
438 Easy
438 of 998
Application +4 | -1
If a,b,ca, b, c are in A.P. and a2,b2,c2a^{2}, b^{2}, c^{2} are in G.P. such that a<b<ca < b < c and a+b+c=34a+b+c=\frac{3}{4}, then the value of aa is
View answer and explanation
Correct option: D
14−122\frac{1}{4}-\frac{1}{2 \sqrt{2}}
Option D (1/4 - 1/(2√2)) is correct because it is the value of 'a' derived by using the properties of A.P. and G.P. and solving the resulting quadratic equation.
439 Easy
439 of 998
Analyzing +4 | -1
If Sn=∑r=1nar=16n(2n2+9n+13),{S_n} = \sum \limits_{r = 1}^n {a_r} = \frac{1}{6}n\left( {2{n^2} + 9n + 13} \right), then ∑r=1nar\sum \limits_{r = 1}^n \sqrt {{a_r}} equals
View answer and explanation
Correct option: C
n(n+3)2\frac{{n\left( {n + 3} \right)}}{2}
Option C (n(n+3)2\frac{{n\left( {n + 3} \right)}}{2}) is correct because it is derived by finding the general term ar{a_r} and then summing the square root of the general term.
440 Easy
440 of 998
Application +4 | -1
If x=1+a+a2+…∞x = 1 + a + {a^2} + \ldots \infty and y=1+b+b2+…∞y = 1 + b + {b^2} + \ldots \infty where aa and bb are proper fractions, then 1+ab+a2b2+…∞1 + ab + {a^2}{b^2} + \ldots \infty equals
View answer and explanation
Correct option: A
xyy+x−1\frac{{xy}}{{y + x - 1}}
Option A (xyy+x−1\frac{{xy}}{{y + x - 1}}) is correct because, given the infinite geometric series for x and y, the expression 1+ab+a2b2+…∞1 + ab + {a^2}{b^2} + \ldots \infty simplifies to xyx+y−1\frac{{xy}}{{x + y - 1}}.
441 Easy
441 of 998
Application +4 | -1
If y=21/log⁡x8,y = {2^{1/{{\log }_x}8}},, then xx equal to
View answer and explanation
Correct option: C
y3{y^3}
Option C is correct because by using the properties of logarithms and simplifying the expression, we can find that x = y³.
442 Easy
442 of 998
Application +4 | -1
If the sum of first nn terms of an AP is cn2c n^2, then the sum of squares of these nn terms is
View answer and explanation
Correct option: C
n(4n2−1)c23\frac{n\left(4 n^2-1\right) c^2}{3}
Option C (n(4n2−1)c23\frac{n(4n^2-1)c^2}{3}) is correct because it is the sum of the squares of the first n terms of the AP, derived from the given sum of the first n terms.
443 Easy
443 of 998
Application +4 | -1
The minimum value of 4x+41−x,x∈R,{4^x} + {4^{1 - x}},x \in R, is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the minimum value is found using the AM-GM inequality, resulting in a minimum value of 4.
444 Medium
444 of 998
Application +4 | -1
If SnS_n denotes the sum of nn terms of an AP, then Sn+3−3Sn+2+3Sn+1−SnS_{n+3}-3 S_{n+2}+3 S_{n+1}-S_n is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because the expression simplifies to zero when using the properties of arithmetic progressions and their sums.
445 Easy
445 of 998
Application +4 | -1
If nn arithmetic means are inserted between aa and 100 such that the ratio of the first mean to the last mean is 1: 7 and a+n=33,a + n = 33, then the value of nn is:
View answer and explanation
Correct option: C
23
Option C (23) is correct because it satisfies the given conditions for the arithmetic means inserted between a and 100.
446 Easy
446 of 998
Analyzing +4 | -1
12−12.22+13.23−14.24+…\frac{1}{2} - \frac{1}{{{{2.2}^2}}} + \frac{1}{{{{3.2}^3}}} - \frac{1}{{{{4.2}^4}}} + \ldots is equal to
View answer and explanation
Correct option: C
log⁡e(32){\log _e}\left( {\frac{3}{2}} \right)
Option C (log⁡e(32){\log _e}(\frac{3}{2})) is correct because the given series is a direct application of the Maclaurin series expansion of log⁡e(1+x)\log_e(1+x) with x=12x = \frac{1}{2}.
447 Medium
447 of 998
Application +4 | -1
The sum of the series 04+0.004+000004+…04+0.004+000004+\ldots is
View answer and explanation
Correct option: C
4099\frac{40}{99}
Option C (4099\frac{{40}}{{99}}) is correct because the sum of the infinite GP is calculated as a/(1-r).
448 Easy
448 of 998
Application +4 | -1
The value of nn for which 704+12(704)+704+\frac{1}{2}(704)+ 14(704)+…\frac{1}{4}(704)+\ldots upto nn terms =1984−12(1984)+14(1984)=1984-\frac{1}{2}(1984)+\frac{1}{4}(1984) −…-\ldots upto nn terms is
View answer and explanation
Correct option: A
5
Option A (5) is correct because solving the equation derived from the given geometric series leads to n=5 as the only valid solution.
449 Easy
449 of 998
Application +4 | -1
Which of the following is the sum of 20 terms of an AP1,4,7,10,…A P 1,4,7,10, \ldots ?
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Correct option: B
590
Option A (590) is correct because the sum of the first 20 terms of the AP 1, 4, 7, 10,... is 590.
450 Medium
450 of 998
Application +4 | -1
If f(n)=[12+n100]f(n)=\left[\frac{1}{2}+\frac{n}{100}\right], where [x][x] denotes the integral part of xx, then the value of ∑n=1100f(n)\sum\limits_{{n=1}}^{100} f(n) is
View answer and explanation
Correct option: B
51
Option B (51) is correct because the sum of the integral parts of the given function from n=1 to 100 is equal to 51.
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