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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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331 Easy
331 of 998
Understanding +4 | -1
If x,y,zx, y, z are in harmonic progression, then which of the following is true?
View answer and explanation
Correct option: D
zx>y\sqrt{z x}>y
Option D (zx>y\sqrt{z x}>y) is correct because in a harmonic progression, the harmonic mean is always less than the geometric mean.
332 Easy
332 of 998
Analyzing +4 | -1
Let Sn=113+1+213+23+…+1+2+…+n13+23+…+n3;n=1,2,3,…{S_n} = \frac{1}{{{1^3}}} + \frac{{1 + 2}}{{{1^3} + {2^3}}} + \ldots + \frac{{1 + 2 + \ldots + n}}{{{1^3} + {2^3} + \ldots + {n^3}}};n = 1,2,3, \ldots
Then, Sn{S_n} is not greater than
View answer and explanation
Correct option: C
2
Option C (2) is correct because the sum of the series is less than 2, as derived by simplifying the terms and using the properties of series.
333 Easy
333 of 998
Analyzing +4 | -1
Let an{a_n} be nnth term of the GP of positive numbers. Let ∑n=1100a2n=α  \sum \limits_{n = 1}^{100} {a_{2n}} = \alpha \; and ∑n=1100a2n=β\sum \limits_{n = 1}^{100} {a_{2n}} = \beta, such that α≠β,\alpha \ne \beta , then the common ratio is
View answer and explanation
Correct option: A
αβ\frac{\alpha }{\beta }
Option A (αβ\frac{\alpha}{\beta}) is correct because the ratio of the sum of even-indexed terms to the sum of odd-indexed terms in a GP is equal to the common ratio.
334 Easy
334 of 998
Analyzing +4 | -1
If ∑k=1n(∑m=1km2)=an4+bn3+cn2+dn+e\sum\limits_{{k=1}}^n\left(\sum\limits_{{m=1}}^k m^2\right)=a n^4+b n^3+c n^2+d n+e, then
View answer and explanation
Correct option: A
a=112a=\frac{1}{12}
Option A (a=112a=\frac{1}{12}) is correct because, after expanding the summation and comparing coefficients, the coefficient of n4n^4 is 112\frac{1}{12}.
335 Easy
335 of 998
Application +4 | -1
The sum of first 20 terms of the sequence 0.7,0.77,0.777,…0.7,0.77,0.777, \ldots, is
View answer and explanation
Correct option: C
781(179+10−20)\frac{7}{81}\left(179+10^{-20}\right)
Option C is correct because the sum of the given series is derived by converting the decimals to fractions and using the formula for the sum of a geometric progression.
336 Easy
336 of 998
Analyzing +4 | -1
If x,y,zx,y,z are ppth,qqth and rrth terms respectively, of and A.P. and also of G.P., then xy−zyz−xzx−y,{x^{y - z}}{y^{z - x}}{z^{x - y}}, is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because when x, y, and z are both in A.P. and G.P., the expression simplifies to 1.
337 Easy
337 of 998
Application +4 | -1
If log⁡102=0.3010…\log _{10} 2=0.3010 \ldots, the number of digits in the number 200020002000^{2000} is
View answer and explanation
Correct option: C
6603
Option C (6603) is correct because the number of digits in 200020002000^{2000} is found by calculating log⁡10(20002000)\log_{10}(2000^{2000}) and adding 1 to the integer part of the result.
338 Easy
338 of 998
Application +4 | -1
x1/2.x1/4.x1/8.x1/16…to∞{x^{1/2}}.{x^{1/4}}.{x^{1/8}}.{x^{1/16}} \ldots \mathrm{to }\infty is equal to
View answer and explanation
Correct option: C
xx
Option C (xx) is correct because the expression simplifies to x by using the properties of exponents and the sum of an infinite geometric series.
339 Easy
339 of 998
Analyzing +4 | -1
If log⁡xa,ax/2{\log _x}a,{a^{x/2}} and log⁡bx{\log _b}x are in G.P., then xxis equal to
View answer and explanation
Correct option: A
log⁡a(log⁡ba){\log _a}\left( {{{\log }_b}a} \right)
Option A (log⁡a(log⁡ba){\log _a}({{\log }_b}a)) is correct because it is the solution for xx when log⁡xa,ax/2{\log _x}a,{a^{x/2}} and log⁡bx{\log _b}x are in G.P.
340 Easy
340 of 998
Application +4 | -1
If (4.2)x=(0.42)y=100,{\left( {4.2} \right)^x} = {\left( {0.42} \right)^y} = 100, then 1x−1y=\frac{1}{x} - \frac{1}{y} =
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (1/2) is correct because by taking logarithms and manipulating the equations, we can find that 1/x - 1/y equals 1/2.
341 Easy
341 of 998
Application +4 | -1
If the sum of the first nn terms of an A.P. is n2n^2, then the difference between two consecutive terms of that A.P. is
View answer and explanation
Correct option: B
2
Option B (2) is correct because the difference between consecutive terms in the A.P. is 2, as derived from the given sum formula.
342 Easy
342 of 998
Analyzing +4 | -1
If Tn=14(n+2)(n+3){T_n} = \frac{1}{4}\left( {n + 2} \right)\left( {n + 3} \right) for n=1,2,3,…,n = 1,2,3, \ldots , then 1T1+1T2+…+1T2003\frac{1}{{{T_1}}} + \frac{1}{{{T_2}}} + \ldots + \frac{1}{{{T_{2003}}}} is equal to
View answer and explanation
Correct option: D
40063009\frac{{4006}}{{3009}}
Option D (40063009\frac{{4006}}{{3009}}) is correct because the sum of the series is 40063009\frac{{4006}}{{3009}}.
343 Easy
343 of 998
Analyzing +4 | -1
If a,ba, b and cc are the three real solutions of the equation
xlog⁡102x+log⁡10x3+3=21x+1−1−1x+1+1x^{\log _{10}^2 x+\log _{10} x^3+3}=\frac{2}{\frac{1}{\sqrt{x+1}-1}-\frac{1}{\sqrt{x+1}+1}}

where, a>b>ca>b>c, then a,b,ca, b, c are in
View answer and explanation
Correct option: B
GP
Option B (GP) is correct because the solutions to the equation form a geometric progression after simplifying and solving for x.
344 Easy
344 of 998
Analyzing +4 | -1
If the sum of the roots of the equation ax2+bx+c=0a{x^2} + bx + c = 0 be equal to the sum of the reciprocals of their squares, then bc2,  ca2,  ab2b{c^2},\;c{a^2},\;a{b^2} will be in
View answer and explanation
Correct option: A
AP
Option C (ac+bd>b2+c2ac + bd > {b^2} + {c^2}) is correct because in a Harmonic Progression, the relationship ac+bd>b2+c2ac + bd > {b^2} + {c^2} holds true.
345 Easy
345 of 998
Application +4 | -1
The fourth, seventh and tenth terms of a G.P. are p,q,rp, q, r respectively, then
View answer and explanation
Correct option: B
q2=prq^2=p r
Option B (q2=prq^2=p r) is correct because in a G.P., if terms are equally spaced, then they are in G.P., which means the square of the middle term equals the product of the other two.
346 Easy
346 of 998
Application +4 | -1
The value of
1+(log⁡en)22  !+(log⁡en)44  !+…is1 + \frac{{{{\left( {{{\log }_e}n} \right)}^2}}}{{2\;!}} + \frac{{{{\left( {{{\log }_e}n} \right)}^4}}}{{4\;!}} + \ldots \mathrm{is}
View answer and explanation
Correct option: C
n+n−12\frac{{n + {n^{ - 1}}}}{2}
Option C (n+n−12\frac{{n + {n^{ - 1}}}}{2}) is correct because the given series is the expansion of elog⁡en+e−log⁡en2\frac{{{e^{{{\log }_e}n}} + {e^{ - {{\log }_e}n}}}}{2}, which simplifies to n+n−12\frac{{n + {n^{ - 1}}}}{2}.
347 Easy
347 of 998
Application +4 | -1
The 5th term of the series 109,13203,23,…\frac{{10}}{9},\frac{1}{3}\sqrt {\frac{{20}}{3}} ,\frac{2}{3}, \ldots is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C (25\frac{2}{5}) is correct because the 5th term is calculated using the formula for the nth term of a Geometric Progression (GP).
348 Easy
348 of 998
Application +4 | -1
If a,b,ca,b,c are in GP and x,yx,y are arithmetic mean of a,ba,b and b,cb,c respectively, then 1x+1y\frac{1}{x} + \frac{1}{y} is equal to
View answer and explanation
Correct option: A
2b\frac{2}{b}
Option A (2/b) is correct because given the conditions, simplifying the expression 1/x + 1/y results in 2/b.
349 Easy
349 of 998
Application +4 | -1
Let a1,a2,a3,…a_{1}, a_{2}, a_{3}, \ldots be aa G. P. such that a1<0,a1+a2=4a_{1}<0, a_{1}+a_{2}=4 and a3+a4=16a_{3}+a_{4}=16. If ∑i=19ai=4λ\sum_{\mathrm{i}=1}^{9} a_{i}=4 \lambda, then λ\lambda is equal to:
View answer and explanation
Correct option: B
-171
Option B (-171) is correct because it is the value of lambda calculated by finding the first term and common ratio of the G.P. and then using the sum formula.
350 Easy
350 of 998
Application +4 | -1
The sum of all two digit natural numbers which leave a remainder 5 when they are divided by 7 is equal to
View answer and explanation
Correct option: B
702
Option B (702) is correct because the sum of the two-digit numbers that leave a remainder of 5 when divided by 7 is 702.
351 Easy
351 of 998
Application +4 | -1
The values of 3log⁡8180+5log⁡2524+7log⁡16153\log \frac{{81}}{{80}} + 5\log \frac{{25}}{{24}} + 7\log \frac{{16}}{{15}} is
View answer and explanation
Correct option: A
log⁡2\log 2
Option A (log⁡2\log 2) is correct because simplifying the logarithmic expression using properties of logarithms results in log⁡2\log 2.
352 Easy
352 of 998
Application +4 | -1
If the sum of the first n terms of a series be 5n2+2n,5{n^2} + 2n, then its second term is
View answer and explanation
Correct option: C
17
Option C (17) is correct because the second term is calculated by subtracting the sum of the first term from the sum of the first two terms.
353 Easy
353 of 998
Application +4 | -1
If three positive numbers a,ba, b and cc are in A.P. such that abc=8\mathrm{abc}=8, then the minimum possible value of bb is :
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Correct option: A
2
Option A (2) is correct because when a=b=c=2, abc=8 and a, b, c are in A.P., thus the minimum possible value of b is 2.
354 Easy
354 of 998
Application +4 | -1
If 11 AM's are inserted between 28 and 10, the number of integral AM's is
View answer and explanation
Correct option: A
5
Option A (5) is correct because by finding the common difference of the AP and substituting it into the general term, we can identify the number of integral arithmetic means.
355 Easy
355 of 998
Application +4 | -1
The sum of 24 terms of the following series 2+8+18+32+...\sqrt 2 + \sqrt 8 + \sqrt {18} + \sqrt {32} + ... is
View answer and explanation
Correct option: B
3002\sqrt 2
Option B is correct because the series can be simplified to an arithmetic progression with a common difference of 2\sqrt{2}, and the sum of the first 24 terms is calculated correctly.
356 Easy
356 of 998
Understanding +4 | -1
1+log⁡22+(log⁡e2)22  !+(log⁡e2)33  !+…1 + {\log _2}2 + \frac{{{{\left( {{{\log }_e}2} \right)}^2}}}{{2\;!}} + \frac{{{{\left( {{{\log }_e}2} \right)}^3}}}{{3\;!}} + \ldots is equal to
View answer and explanation
Correct option: A
22
Option A (2) is correct because the given series is the expansion of exe^x where x is loge2log_e 2.
357 Easy
357 of 998
Analyzing +4 | -1
The sum to the infinity of the series 1+23+632+1033+1434+  …1 + \frac{2}{3} + \frac{6}{{{3^2}}} + \frac{{10}}{{{3^3}}} + \frac{{14}}{{{3^4}}} + \; \ldots is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the sum to infinity of the given series can be found by manipulating the series and using the formula for the sum of an infinite geometric progression.
358 Easy
358 of 998
Application +4 | -1
If the sums of nn terms of two A.P.'s are in the ratio (7n+1):(4n+27)(7 n+1):(4 n+27), then the ratio of their nth n^{\text {th }} terms is
View answer and explanation
Correct option: B
(14n−6):(8n+23)(14 n-6):(8 n+23)
Option B ((14n−6):(8n+23)(14n-6):(8n+23)) is correct because the ratio of the nth terms is obtained by replacing n by (2n-1) in the given ratio of the sums.
359 Easy
359 of 998
Understanding +4 | -1
Assertion: Sum to nn terms of the geometric progression
x3,x5,x7,…(x≠±1) is x3(1−x2n)(1−x2).x^3, x^5, x^7, \ldots(x \neq \pm 1) \text { is } \frac{x^3\left(1-x^{2 n}\right)}{\left(1-x^2\right)} .

Reason: If 'aa' is the first term and r{r} is common ratio of a G.P. then sum to n{n} terms is given as
Sn=a(rn−1)r−1 or =a(1−rn)1−r if r≠1.{S}_{{n}}=\frac{{a}\left({r}^{{n}}-1\right)}{{r}-1} \text { or }=\frac{{a}\left(1-{r}^{{n}}\right)}{1-{r}} \text { if } {r} \neq 1 .
View answer and explanation
Correct option: A
Assertion is correct, reason is correct; reason is a correct explanation for assertion
Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion) is correct because the sum to n terms of the given G.P. is correctly calculated using the formula provided in the reason.
360 Easy
360 of 998
Application +4 | -1
Let a1, a2,..., an{a_1},\,{a_2},...,\,{a_n} be a given AP. Whose common difference is an integer and Sn =a1 + a2+...+an.{\mathrm{S}_\mathrm{n}}\, = {a_{1\,}} + \,{a_2} + ... + {a_n}. If a1 = 1,  an = 300{a_1}\, = \,1,\,\,{a_n}\, = \,300 and 15≤n≤50,15 \le n \le 50, then the ordered pair (Sn−4, an−4)\left( {{S_{n - 4}},\,{a_{n - 4}}} \right) is equal to
View answer and explanation
Correct option: D
(2490,  248)\left( {2490,\,\,248} \right)
Option D ((2490, 248)) is correct because it calculates the sum and the nth term of the arithmetic progression after determining the common difference and the number of terms.
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