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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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Showing 511-540 of 998 questions

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511 Easy
511 of 998
Application +4 | -1
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is 2719\frac{{27}}{{19}} Then, the common ration of this series is
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{2}{3}) is correct because it's the common ratio of the geometric series that satisfies both the sum and the sum of cubes conditions.
512 Easy
512 of 998
Application +4 | -1
The harmonic mean of 4,8 and 16 is
View answer and explanation
Correct option: C
6.85
Option C (6.85) is correct because it is the result of calculating the harmonic mean of the given numbers.
513 Easy
513 of 998
Application +4 | -1
Sum to nn terms of the series 1+(1+2)+(1+2+1+(1+2)+(1+2+ 3) +…+\ldots is
View answer and explanation
Correct option: D
16n(n+1)(n+2)\frac{1}{6} n(n+1)(n+2)
Option D (16n(n+1)(n+2)\frac{1}{6} n(n+1)(n+2)) is correct because it represents the sum to n terms of the given series, derived by summing the individual terms and simplifying.
514 Easy
514 of 998
Application +4 | -1
The value of

(1+a2x22  !+a4x44  !+…)2−(ax+a3x33  !+a5x55  !+…)2,  is{\left( {1 + \frac{{{a^2}{x^2}}}{{2\;!}} + \frac{{{a^4}{x^4}}}{{4\;!}} + \ldots } \right)^2} - {\left( {ax + \frac{{{a^3}{x^3}}}{{3\;!}} + \frac{{{a^5}{x^5}}}{{5\;!}} + \ldots } \right)^2},\;\mathrm{is}
View answer and explanation
Correct option: D
1
Option D (1) is correct because the expression simplifies to 1 using the exponential identities.
515 Easy
515 of 998
Application +4 | -1
The sum of the first 20 terms common between the series 3 +7+11+15+………+7+11+15+\ldots \ldots \ldots and 1+6+11+16+……1+6+11+16+\ldots \ldots, is
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Correct option: D
4020
Option D (4020) is correct because it is the sum of the first 20 common terms of the given series, which form an arithmetic progression.
516 Easy
516 of 998
Analyzing +4 | -1
If a,  b,  c,  da,\;b,\;c,\;d are any four consecutive coefficients of any expanded binomial, then a+ba,b+cb,c+dc\frac{{a + b}}{a},\frac{{b + c}}{b},\frac{{c + d}}{c} are in
View answer and explanation
Correct option: C
HP
Option C (HP) is correct because the given expressions form a Harmonic Progression when a, b, c, and d are consecutive binomial coefficients.
517 Easy
517 of 998
Analyzing +4 | -1
Sum of nn terms to the series 5+7+13+31+85+……5+7+13+31+85+\ldots \ldots. , is
View answer and explanation
Correct option: A
4n+12(3n−1)4 n+\frac{1}{2}\left(3^n-1\right)
Option A (4n+12(3n−1)4 n+\frac{1}{2}\left(3^n-1\right)) is correct because it represents the sum of the given series, derived by finding the nth term and summing the series.
518 Easy
518 of 998
Analyzing +4 | -1
12.5+15.8+18.11+...upto  n\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + ...\mathrm{upto}\;n terms is equal to
View answer and explanation
Correct option: C
n6n+4\frac{n}{{6n + 4}}
Option C (n6n+4\frac{n}{{6n + 4}}) is correct because it represents the sum of the series up to nn terms.
519 Easy
519 of 998
Application +4 | -1
If log⁡2(5.2x+1),log⁡4(21−x+1)\log _2\left(5.2^x+1\right), \log _4\left(2^{1-x}+1\right) and 1 are in A.P, then xx equals
View answer and explanation
Correct option: B
1−log⁡251-\log _2 5
Option B (1−log⁡251-\log _2 5) is correct because by substituting y=2xy = 2^x and solving the resulting quadratic equation, we find x=1−log⁡25x = 1 - \log_2 5.
520 Easy
520 of 998
Application +4 | -1
The value of x=21/441/881/16161/32…x=2^{1 / 4} 4^{1 / 8} 8^{1 / 16} 16^{1 / 32} \ldots is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the infinite product can be expressed as 2 raised to the power of an infinite sum, which converges to 1.
521 Easy
521 of 998
Analyzing +4 | -1
The coefficient of xn−2x^{n-2} in the polynomial (x−1)(x−2)(x−3)…(x−n)(x-1)(x-2)(x-3) \ldots(x-n), is
View answer and explanation
Correct option: B
n(n2−1)(3n+2)24\frac{n\left(n^2-1\right)(3 n+2)}{24}
Option B (n(n2−1)(3n+2)24\frac{{n(n^2-1)(3n+2)}}{{24}}) is correct because it represents the coefficient of xn−2x^{n-2} in the polynomial, calculated by summing the product of the roots taken two at a time.
522 Easy
522 of 998
Analyzing +4 | -1
If x+y1−xy,y,y+z1−yz\frac{{x + y}}{{1 - xy}},y,\frac{{y + z}}{{1 - yz}} be in A.P., then x,1y,zx,\frac{1}{y},z will be in
View answer and explanation
Correct option: C
H.P.
Option C (H.P.) is correct because the given condition leads to the conclusion that x,1y,zx, \frac{1}{y}, z are in Harmonic Progression.
523 Easy
523 of 998
Application +4 | -1
If 12+22+32+…+20092=(2009)(4019)(335)1^2+2^2+3^2+\ldots+2009^2= (2009) (4019) (335) and

(1)(2009)+(2)(2008)+(3)(2007)+…+(2009)(1)=(2009)(335)(x)(1) (2009) + (2) (2008) + (3) (2007) +\ldots+(2009)(1)=(2009)(335)(x),

then xx equals
View answer and explanation
Correct option: A
2011
Option A (2011) is correct because by adding the two given series and using the formula for the sum of squares and the sum of integers, we can solve for x.
524 Easy
524 of 998
Application +4 | -1
Number of value(s) of 'xx' satisfying the equation log⁡2(log⁡3(x2))=1\log _2\left(\log _3\left(x^2\right)\right)=1 is/are:
View answer and explanation
Correct option: C
2
Option C is correct because solving the logarithmic equation leads to two solutions: x = 3 and x = -3.
525 Easy
525 of 998
Application +4 | -1
If a,b,ca,b,c are in H.P., then the value of b+ab−a+b+cb−c\frac{{b + a}}{{b - a}} + \frac{{b + c}}{{b - c}} is
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Correct option: B
2
Option B (2) is correct because using the property of Harmonic Progression and applying componendo dividendo simplifies the expression to 2.
526 Easy
526 of 998
Application +4 | -1
If sin⁡θ,2(sin⁡θ+1),6sin⁡θ+6\sin \theta, \sqrt{2}(\sin \theta+1), 6 \sin \theta+6 are in GP\mathrm{GP}, then the fifth term is
View answer and explanation
Correct option: C
162
Option C (162) is correct because by finding the common ratio and first term, we can calculate the fifth term of the GP.
527 Easy
527 of 998
Application +4 | -1
The sum of 11 terms of an A.P. whose middle term is 30, is
View answer and explanation
Correct option: B
330
Option B (330) is correct because the sum of an arithmetic progression is equal to the number of terms multiplied by the middle term.
528 Easy
528 of 998
Application +4 | -1
Let f(n)=[15+3n100]nf(n)=\left[\frac{1}{5}+\frac{3 n}{100}\right] n, where [x][x] denotes the greatest integer less than or equal to xx. Then
∑n=161f(n)\sum\limits_{{n=1}}^{61} f(n)
View answer and explanation
Correct option: D
1661
Option D (1661) is correct because by evaluating the function for different ranges of n, we can simplify the summation and calculate the result.
529 Easy
529 of 998
Application +4 | -1
If 12+22+…+n2=10151^2+2^2+\ldots+n^2=1015, then value of nn is
View answer and explanation
Correct option: B
14
Option B (14) is correct. By using the formula for the sum of squares and solving the resulting equation, we find n = 14.
530 Easy
530 of 998
Analyzing +4 | -1
If x=∑n=0∞an,y=∑n=0∞bn,z=∑n=0∞cn,x = \sum\limits_{n = 0}^\infty {{a^n}} ,y = \sum\limits_{n = 0}^\infty {{b^n}} ,z = \sum\limits_{n = 0}^\infty {{c^n}} , where a,b,ca,b,c are in A.P and ∣a∣<1,∣b∣<1,∣c∣<1,  abc≠0,\left| a \right| < 1,\left| b \right| < 1,\left| c \right| < 1,\,\,abc \ne 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 if a, b, c are in A.P., then their reciprocals with a constant subtracted are in Harmonic Progression, implying x, y, z are in H.P., and therefore their reciprocals are in A.P.
531 Easy
531 of 998
Understanding +4 | -1
If {an}\left\{ {{a_n}} \right\} is a sequence with a0=p{a_0} = p and an−an−1=ran−1{a_n} - {a_{n - 1}} = r{a_{n - 1}} for n≥1n \ge 1, then the terms of the sequence are in
View answer and explanation
Correct option: B
A geometric progression
Option B (A geometric progression) is correct because the recursive relation implies a constant ratio between consecutive terms.
532 Easy
532 of 998
Application +4 | -1
For a sequence <an>,a1=5< a_n > , a_1=5, and ar+1ar=12 ∀ r\frac{a_{r+1}}{a_r}=\frac{1}{2} \, \forall \, r, then ∑n=1∞a2n−1\sum\limits_{{n=1}}^{\infty} a_{2 n-1} is
View answer and explanation
Correct option: C
20/320 / 3
Option C (20/320 / 3) is correct because the sum of the specified infinite geometric series is calculated correctly.
533 Easy
533 of 998
Application +4 | -1
nn arithmetic means are inserted between 3 and 17. If the ratio of first to last mean is 1 : 3, then nn is
equal to:
View answer and explanation
Correct option: C
6
Option C (6) is correct because by using the arithmetic means formula and the given ratio, we can solve for the number of arithmetic means.
534 Easy
534 of 998
Application +4 | -1
150 workers were engaged to finish a piece of work in a certain number of days. 4 workers dropped the second day, 4 more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is
View answer and explanation
Correct option: C
25
Option C (25) is correct because it is the number of days calculated by setting up an AP series representing the decreasing number of workers and solving for the original number of days.
535 Easy
535 of 998
Analyzing +4 | -1
∑r=1nr2−∑m=1n∑r=1mr\sum\limits_{{r=1}}^n r^2-\sum\limits_{{m=1}}^n \sum\limits_{{r=1}}^m r is equal to
View answer and explanation
Correct option: C
12{∑r=1nr2−∑r=1nr}\frac{1}{2}\left\{\sum\limits_{{r=1}}^n r^2-\sum\limits_{{r=1}}^n r\right\}
Option C (12{∑r=1nr2−∑r=1nr}\frac{1}{2}\{\sum\limits_{{r=1}}^n r^2-\sum\limits_{{r=1}}^n r\}) is correct because it is the simplified form of the given expression after evaluating the inner summation.
536 Easy
536 of 998
Analyzing +4 | -1
The sum of the series 12⋅  221  !+22⋅  322  !+32⋅  423  !+…,\frac{{{1^2} \cdot \;{2^2}}}{{1\;!}} + \frac{{{2^2} \cdot \;{3^2}}}{{2\;!}} + \frac{{{3^2} \cdot \;{4^2}}}{{3\;!}} + \ldots , is
View answer and explanation
Correct option: A
27e27e
Option A (27e27e) is correct because the series can be decomposed into sums involving ee, resulting in 27e27e.
537 Easy
537 of 998
Application +4 | -1
Solution of 0<∣3x+1∣<130 < |3 x+1| < \frac{1}{3} is
View answer and explanation
Correct option: C
(−4/9,−2/9)−{−1/3}(-4 / 9,-2 / 9)-\{-1 / 3\}
Option C ((-4 / 9,-2 / 9)-{-1 / 3}) is correct because it represents the interval where the absolute value condition is met, excluding the point where the absolute value is zero.
538 Easy
538 of 998
Analyzing +4 | -1
∑k=1∞1k!(∑n=1∞2n−1)\sum \limits_{k = 1}^\infty \frac{1}{{k!}}\left( {\sum \limits_{n = 1}^\infty {2^{n - 1}}} \right) is equal to
View answer and explanation
Correct option: D
e2−e{e^2} - e
Option D (e2−e{e^2} - e) is correct because it is derived by splitting the sum into two separate exponential series and evaluating them.
539 Easy
539 of 998
Application +4 | -1
If 2(y−a)2\left( {y - a} \right) is the H.M. between y−xy - x and y−z,y - z, then x−a,  y−a,  z−ax - a,\;y - a,\;z - aare in
View answer and explanation
Correct option: B
G.P.
The correct answer is B (G.P.) because the given harmonic mean condition leads to the conclusion that x-a, y-a, and z-a are in geometric progression.
540 Easy
540 of 998
Application +4 | -1
The second term of an AP is (x−y)(x-y) and the fifth term is (x+y)(x+y), then its first term is
View answer and explanation
Correct option: D
x−53yx-\frac{5}{3} y
Option D (x−53yx-\frac{5}{3} y) is correct because by using the information about the second and fifth terms of the AP, we can find the first term using the properties of arithmetic progressions.
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