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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Mixed difficulty

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391 Easy
391 of 998
Application +4 | -1
The ratio of the sum of first three terms of a G.P. to the sum of first 6 terms of the G.P. is 125:152125: 152, then common ratio of the G.P. is
View answer and explanation
Correct option: C
35\frac{3}{5}
Option C (3/5) is correct because, given the ratio of the sums, the common ratio of the G.P. is 3/5.
392 Medium
392 of 998
Application +4 | -1
If p,q,rp, q, r are in APA P and x,y,zx, y, z are in GP, then xq−r⋅yr−p⋅zp−qx^{q-r} \cdot y^{r-p} \cdot z^{p-q} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because when p, q, r are in AP and x, y, z are in GP, the expression simplifies to 1.
393 Easy
393 of 998
Application +4 | -1
Let Sn{S_n} be the sum of the first nn terms of an arithmetic progression, If S3n=3S2n{S_{3n}} = 3{S_{2n}}, then the value of S4nS2n\frac{{{S_{4n}}}}{{{S_{2n}}}} is
View answer and explanation
Correct option: A
6
Option A (6) is correct because it is the value of the ratio of the sums, calculated using the properties of arithmetic progressions.
394 Medium
394 of 998
Application +4 | -1
In a GP, if (m+n)(m+n) th term is pp and (m−n)(m-n) th term is qq, then its mm th term is
View answer and explanation
Correct option: C
pq\sqrt{p q}
Option C (pq\sqrt{p q}) is correct because the mth term of the GP is found by multiplying the (m+n)th and (m-n)th terms and taking the square root.
395 Easy
395 of 998
Application +4 | -1
Let a1,a2,a3……….a_{1}, a_{2}, a_{3} \ldots \ldots \ldots .. be terms on A.P. If a1+a2+………..apa1+a2+………..+aq=p2q2,p≠q\frac{a_{1}+a_{2}+\ldots \ldots \ldots . . a_{p}}{a_{1}+a_{2}+\ldots \ldots \ldots . .+a_{q}}=\frac{p^{2}}{q^{2}}, p \neq q, then a6a21\frac{a_{6}}{a_{21}} equals
View answer and explanation
Correct option: D
1141\frac{11}{41}
Option D (11/41) is correct because it is obtained by substituting appropriate values into the formula for the sum of an arithmetic progression and simplifying.
396 Easy
396 of 998
Application +4 | -1
If a1,a2,…,ana_1, a_2, \ldots, a_n are in HP\mathrm{HP}, then the expression a1a2+a2a3+…+an−1ana_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n is equal to
View answer and explanation
Correct option: D
(n−1)a1an(n-1) a_1 a_n
Option D ((n−1)a1an(n-1) a_1 a_n) is correct because the sum of the product of consecutive terms in a harmonic progression can be simplified to this expression.
397 Medium
397 of 998
Application +4 | -1
The sum of the series (1+2)+(1+2+22)+(1+2+22+23)+...upto  n\left( {1 + 2} \right) + \left( {1 + 2 + {2^2}} \right) + \left( {1 + 2 + {2^2} + {2^3}} \right) + ...\mathrm{upto}\;n terms is
View answer and explanation
Correct option: A
2n+2−n−4{2^{n + 2}} - n - 4
The correct answer is A (2n+2−n−4{2^{n + 2}} - n - 4) because summing the series of geometric progressions and simplifying leads to this result.
398 Easy
398 of 998
Application +4 | -1
If H1,H2{H_1},{H_2} are two harmonic means between two positive numbers aa and b  (a≠b)b\;\left( {a \ne b} \right), AA and GG are the arithmetic and geometric means between aa and bb, then H2+H1H2H1\frac{{{H_2} + {H_1}}}{{{H_2}{H_1}}} is
View answer and explanation
Correct option: D
2AG2\frac{{2A}}{{{G^2}}}
Option D (2AG2\frac{{2A}}{{{G^2}}}) is correct because it is derived by simplifying the expression using the definitions of harmonic, arithmetic, and geometric means.
399 Easy
399 of 998
Application +4 | -1
Let a,b∈R+a, b \in R^{+}, such that log⁡27a+log⁡9b=72\log _{27} a+\log _9 b=\frac{7}{2} and log⁡27b+log⁡9a=23\log _{27} b+\log _9 a=\frac{2}{3}, then aba b is equal to:
View answer and explanation
Correct option: B
243
Option B (243) is correct because by converting the logarithms to a common base and solving the system of equations, we find that ab=243ab = 243.
400 Medium
400 of 998
Understanding +4 | -1
The coefficients of xn{x^n} in the expansion of log⁡a(1+x){\log _a}\left( {1 + x} \right) is
View answer and explanation
Correct option: B
(−1)n−1nlog⁡ae\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{\log _a}e
Option B ((−1)n−1nlog⁡ae\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{n}{\log _a}e) is correct because it is the coefficient of xnx^n in the expansion of log⁡a(1+x){\log _a}(1 + x).
401 Easy
401 of 998
Application +4 | -1
Sum of the series S=12−22+32−42+S=1^2-2^2+3^2-4^2+ …−20082+20092\ldots-2008^2+2009^2 is
View answer and explanation
Correct option: A
2019045
Option A (2019045) is correct because the sum of the series can be simplified by pairing terms and using the formula for the sum of the first n integers.
402 Easy
402 of 998
Application +4 | -1
The sum of the intercepts cut off by the axes on the lines x+y=a,x+y=ar,x+y=ar2,…x+y=a, x+y=a r, x+y=a r^2, \ldots where a≠0a \neq 0 and r=1/2r=1 / 2 is
View answer and explanation
Correct option: C
22a2 \sqrt{2} a
The correct answer is 22a2 \sqrt{{2}} a because the sum of intercepts forms a geometric series with a common ratio of 1/2, and the sum to infinity is calculated.
403 Medium
403 of 998
Application +4 | -1
If x,y,zx,y,z are positive integers, then (x+y)(y+z)(z+x)\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right) is
View answer and explanation
Correct option: B
>8  xyz> 8\;xyz
Option B (>8  xyz> 8\;xyz) is correct because it follows from applying the AM-GM inequality to the pairs (x, y), (y, z), and (z, x).
404 Easy
404 of 998
Analyzing +4 | -1
The sum of the series 2{7−1+3−1⋅7−3+5−1⋅7−5+…}2\left\{ {{7^{ - 1}} + {3^{ - 1}} \cdot {7^{ - 3}} + {5^{ - 1}} \cdot {7^{ - 5}} + \ldots } \right\} is
View answer and explanation
Correct option: A
log⁡e(43){\log _e}\left( {\frac{4}{3}} \right)
Option A (log⁡e(43){\log _e}(\frac{4}{3})) is correct because the series can be recognized as a Taylor series expansion of a logarithmic function.
405 Easy
405 of 998
Analyzing +4 | -1
If x=1+log⁡abc,y=1+log⁡bca,z=1+log⁡cab,x=1+\log _a b c, y=1+\log _b c a, z=1+\log _c a b, then xyzxy+yz+zx\frac{x y z}{x y+y z+z x} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because by using logarithmic properties, we can simplify the expression and find that the given fraction equals 1.
406 Easy
406 of 998
Application +4 | -1
The sum of i−2−3i+4….i - 2 - 3i + 4 \ldots . upto 100 terms, where i=−1i = \sqrt { - 1} is
View answer and explanation
Correct option: A
50(1−i)50\left( {1 - i} \right)
Option A (50(1−i)50\left( {1 - i} \right)) is correct because it is the result of summing the series by manipulating and simplifying the expression using properties of complex numbers.
407 Medium
407 of 998
Analyzing +4 | -1
The sixth term of an A.P. is equal to 2 the value of the common difference of the A.P. which makes the product a1a4a5{a_1}{a_4}{a_5} least is given by
View answer and explanation
Correct option: C
x=2/3x = 2/3
Option C (x=2/3x = 2/3) is correct because it minimizes the product a1a4a5{a_1}{a_4}{a_5} by finding the critical points and checking the second derivative.
408 Easy
408 of 998
Application +4 | -1
Suppose a1,a2,…a_1, a_2, \ldots are in A.P. If a8:a5=a_8: a_5= 3:23: 2, then a17:a23a_{17}: a_{23} is:
View answer and explanation
Correct option: B
3:43: 4
Option B (3:4) is correct because, given the ratio of the 8th and 5th terms, we can find the relationship between the first term and common difference, then find the ratio of the 17th and 23rd terms.
409 Easy
409 of 998
Application +4 | -1
A species has an initial population 4104^{10}. At the end of first day, the population increases by 50%50 \%. At the end of second day, it decreases by the same percentage. If the process continues in the same pattern, the number of days for the population to reach 3103^{10} is
View answer and explanation
Correct option: B
20
Option B (20) is correct because after 20 days, with alternating 50% increases and decreases, the population reaches 310.
410 Medium
410 of 998
Application +4 | -1
The harmonic mean of two numbers is 4 and the arithmetic and geometric means satisfy the relation 2  A+G2=27,2\;A + {G^2} = 27, the numbers are
View answer and explanation
Correct option: A
6, 3
Option A (6, 3) is correct because the harmonic mean is 4, and the arithmetic and geometric means satisfy the given relation, leading to the numbers 6 and 3.
411 Easy
411 of 998
Application +4 | -1
If AM of two terms is 9 and HM is 36 , then GMG M will be
View answer and explanation
Correct option: A
18
Option A (18) is correct because the Geometric Mean (GM) is the square root of the product of the Arithmetic Mean (AM) and Harmonic Mean (HM).
412 Easy
412 of 998
Application +4 | -1
The sum of the series 1⋅2⋅3+2⋅3⋅4+3⋅4⋅5+...1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + ... to nn terms is
View answer and explanation
Correct option: C
14n(n+1)(n+2)(n+3)\frac{1}{4}n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)
Option C (14n(n+1)(n+2)(n+3)\frac{1}{4}n(n + 1)(n + 2)(n + 3)) is correct because it represents the sum of the series 1⋅2⋅3+2⋅3⋅4+3⋅4⋅5+...1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + ... to nn terms, which can be found by expressing the general term as n(n+1)(n+2)n(n+1)(n+2) and summing it.
413 Easy
413 of 998
Application +4 | -1
If 112+122+132+⋯\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots upto ∞=π26\infty=\frac{\pi^2}{6}, then value of 112+132+152+⋯\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots upto ∞\infty is
View answer and explanation
Correct option: C
π28\frac{\pi^2}{8}
Option C (π28\frac{\pi^2}{8}) is correct because it is derived by subtracting the sum of even-numbered terms from the total sum of the series.
414 Easy
414 of 998
Application +4 | -1
The AM, HM and GM between two numbers are 14415,  15\frac{{144}}{{15}},\;15 and 12, but not necessarily in this order. Then, HM, GM and AM respectively are
View answer and explanation
Correct option: B
14415,12,  15\frac{{144}}{{15}},12,\;15
Option A is correct because the GM is always between the AM and HM. Given AM and GM, HM can be calculated and matched with the given values.
415 Easy
415 of 998
Application +4 | -1
nnth term of the series 1+45+752+1053+...1 + \frac{4}{5} + \frac{7}{{{5^2}}} + \frac{{10}}{{{5^3}}} + ... will be
View answer and explanation
Correct option: C
3n−25n−1\frac{{3n - 2}}{{{5^{n - 1}}}}
Option C (3n−25n−1\frac{{3n - 2}}{{{{5^{n - 1}}}}}) is correct because the nth term of the series is derived from the corresponding Arithmetic Progression (AP) and Geometric Progression (GP).
416 Medium
416 of 998
Application +4 | -1
The sum of the series : 2+2(d)2+3(6)2+…^{2}+2(d)^{2}+3(6)^{2}+\ldots upto 10 terms is :
View answer and explanation
Correct option: C
12100
Option C (12100) is correct because the series can be simplified to a multiple of the sum of cubes of the first 10 natural numbers.
417 Easy
417 of 998
Application +4 | -1
If the sum to nn terms of an A.P. is 3n2+5n3 n^2+5 n, while Tm=164T_m=164, then value of mm is
View answer and explanation
Correct option: C
27
Option C (27) is correct because solving the equation Tm=Sm−Sm−1=164T_m = S_m - S_{m-1} = 164 yields m=27m = 27.
418 Easy
418 of 998
Application +4 | -1
If a2+4b2=12ab,{a^2} + 4{b^2} = 12ab, then log⁡(a+2b)=\log \left( {a + 2b} \right) =
View answer and explanation
Correct option: C
12(log⁡a+log⁡b+4log⁡2)\frac{1}{2}\left( {\log a + \log b + 4\log 2} \right)
Option C (12(log⁡a+log⁡b+4log⁡2)\frac{1}{2}\left( {\log a + \log b + 4\log 2} \right)) is correct because by completing the square and applying logarithm properties, we can derive the given expression.
419 Easy
419 of 998
Application +4 | -1
Let a,ba, b and c form a GP of common ratior, with 0<r<10 < r < 1. If a,2ba, 2 b and 3c3 c form an APA P, then rr equals
View answer and explanation
Correct option: B
1/31 / 3
Option B (1/3) is correct because solving the equation derived from the A.P. condition yields r=1/3.
420 Easy
420 of 998
Application +4 | -1
The sum of the first ten terms of an AP\mathrm{AP} is four times the sum of the first five terms, the ratio of the first term to the common difference is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option B ((n−1)r1−r2\frac{(n-1) r}{1-r^2}) is correct because it represents the sum of the given series where the terms are in GP, derived by simplifying the general term and summing.
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