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JEE • MATHEMATICS • SEQUENCE OF SERIES

Sequence of series practice · Medium

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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 Medium
331 of 564
Application +4 | -1
The sum of numbers of three digits which are divisible by 7 , is
View answer and explanation
Correct option: A
70336
Option A (70336) is correct because it is the sum of all three-digit numbers divisible by 7, calculated using the arithmetic series formula.
332 Medium
332 of 564
Application +4 | -1
Let aa and bb be two positive real numbers. Suppose A1A_1, A2A_2 are two arithmetic means; G1,G2G_1, G_2 are two geometric means and H1,H2{H}_1, {H}_2 are two harmonic means between aa and bb, then A1+A2H1+H2\frac{A_1+A_2}{H_1+H_2} is equal to
View answer and explanation
Correct option: A
2(a2+b2)+5ab9ab\frac{2\left(a^2+b^2\right)+5 a b}{9 a b}
Option A is correct because the expression can be simplified using the formulas for arithmetic, geometric, and harmonic means.
333 Medium
333 of 564
Application +4 | -1
A person purchases 1 kg1 \mathrm{~kg} tomatoes from each of the 4 places at the rate of 1 kg,2 kg,3 kg1 \mathrm{~kg}, 2 \mathrm{~kg}, 3 \mathrm{~kg} and 4 kg4 \mathrm{~kg} per rupee respectively. On the average, he has purchased x kgx \mathrm{~kg} tomatoes per rupee, then the value of xx is
View answer and explanation
Correct option: C
1.92
Option C (1.92) is correct because the average rate is best represented by the Harmonic Mean when dealing with rates per unit.
334 Medium
334 of 564
Analyzing +4 | -1
The value of lim⁡r→−∞∑r=1ntan⁡−1(12r2)\lim _{r \rightarrow-\infty} \sum_{r=1}^n \tan ^{-1}\left(\frac{1}{2 r^2}\right) is
View answer and explanation
Correct option: B
π4\frac{\pi}{4}
Option B (π4\frac{\pi}{4}) is correct because the limit of the sum of arctangents can be simplified using the arctangent subtraction formula and telescoping series.
335 Medium
335 of 564
Analyzing +4 | -1
If a,a1,a2,…,a2n,ba,{a_1},{a_2}, \ldots ,{a_{2n}},b are in arithmetic progression and a,g1,g2,…,g2n,ba,{g_1},{g_2}, \ldots ,{g_{2n}},b are in geometric progression and hh is the harmonic mean of aa and bb, then a1+a2ng1g2n+a2+a2n−1g2g2n−1+...+an+an+1gngn+1\frac{{{a_1} + {a_{2n}}}}{{{g_1}{g_{2n}}}} + \frac{{{a_2} + {a_{2n - 1}}}}{{{g_2}{g_{2n - 1}}}} + ... + \frac{{{a_n} + {a_{n + 1}}}}{{{g_n}{g_{n + 1}}}} is equal to
View answer and explanation
Correct option: D
2nh\frac{{2n}}{h}
Option D (2nh\frac{{2n}}{h}) is correct because by properties of AP and GP, the expression simplifies to a constant term involving the harmonic mean.
336 Medium
336 of 564
Application +4 | -1
A man wants to cut three lengths from a single piece of board of length 91 cm91 \mathrm{~cm}. The second length is to be 3 cm3 \mathrm{~cm} longer than the shortest and the third length is to be twice as long as the shortest. The possible length of the shortest board, if the third piece is to be at least 5 cm5 \mathrm{~cm} longer than the second, is
View answer and explanation
Correct option: B
greater than or equal to 8 cm8 \mathrm{~cm} but less than or equal to 22 cm22 \mathrm{~cm}
The correct answer is B (greater than or equal to 8 cm but less than or equal to 22 cm) because the length of the shortest board must satisfy two inequalities derived from the given conditions.
337 Medium
337 of 564
Application +4 | -1
In a certain AP, 5 times the 5 th term is equal to 8 times the 8 th term, then 13 th term is
View answer and explanation
Correct option: D
0
Option D (0) is correct because if 5 times the 5th term equals 8 times the 8th term in an AP, the 13th term must be 0.
338 Medium
338 of 564
Analyzing +4 | -1
The value of ∑k=1n(n−k)cos⁡2kπn (n≥3)\sum\limits_{{k=1}}^n(n-k) \cos \frac{2 k \pi}{n} \, (n \geq 3) is
View answer and explanation
Correct option: D
−n-n
Option D is correct because the sum can be simplified using trigonometric identities and the formula for the sum of cosines in an arithmetic progression.
339 Medium
339 of 564
Application +4 | -1
Let α,β\alpha, \beta and γ\gamma be three positive real numbers. Let f(x)=αx5+βx3+γx,x∈Rf(x)=\alpha x^5+\beta x^3+\gamma x, x \in \mathbb{R} and g:R→Rg: \mathbb{R} \rightarrow \mathbb{R} be such that g(f(x))=xg(f(x))=x for all x∈Rx \in \mathbb{R}. If a1,a2,a3,…,ana_1, a_2, a_3, \ldots, a_n be in arithmetic progression mean is zero, then the value of f(g(1n∑i=1nf(ai)))f\left(g\left(\frac{1}{n} \sum\limits_{i=1}^n f\left(a_i\right)\right)\right) is equal to
View answer and explanation
Correct option: A
0
The correct answer is A (0) because the function f(x)f(x) is odd, and the sum of an arithmetic progression with a mean of zero is zero, leading to f(g(0))=0f(g(0)) = 0.
340 Medium
340 of 564
Application +4 | -1
An infinite GP has the first term ′x′'x' and sum 5, then xx belongs to
View answer and explanation
Correct option: C
0<x<100 < x < 10
Option C is correct because the condition for the sum of an infinite GP to exist requires the common ratio to be between -1 and 1, leading to the inequality 0 < x < 10.
341 Medium
341 of 564
Application +4 | -1
the sum of nn terms of the infinite series

1⋅32+2⋅52+3⋅72+...∞1 \cdot {3^2} + 2 \cdot {5^2} + 3 \cdot {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 (n6\frac{n}{6}\left( {n + 1} \right)\left( {6{n2} + 14n + 7} \right)) is correct because it is the correct formula for the sum of the given arithmetico-geometric series.
342 Medium
342 of 564
Application +4 | -1
If Sn=nP+n2(n−1)QS_n=n P+\frac{n}{2}(n-1) Q, where SnS_n denotes the sum of first nn terms of an AP, then the common difference is
View answer and explanation
Correct option: D
QQ
Option D (QQ) is correct because the common difference in an AP is the coefficient of the n2(n−1)\frac{{n}}{{2}}(n-1) term in the sum formula.
343 Medium
343 of 564
Application +4 | -1
Given two numbers aa and bb. Let AA denote the single A.M. and SS denote the sum of nn A.M.’s between aa and bb, then S/AS/A depends on
View answer and explanation
Correct option: D
nn
Option D (nn) is correct because the ratio S/A simplifies to n, independent of a and b.
344 Medium
344 of 564
Application +4 | -1
If x>0x>0 and log⁡2x+log⁡2(x)+log⁡2(x4)\log _2 x+\log _2(\sqrt{x})+\log _2(\sqrt[4]{x}) +log⁡2(x8)+log⁡2(x16)+…=4+\log _2(\sqrt[8]{x})+\log _2(\sqrt[16]{x})+\ldots=4, then xx equals
View answer and explanation
Correct option: C
4
Option C (4) is correct because simplifying the logarithmic expression and solving for x gives x = 4.
345 Medium
345 of 564
Application +4 | -1
Let a1,a2,…,a10{a_1},{a_2}, \ldots ,{a_{10}} be in AP and h1,h2,…,h10{h_1},{h_2}, \ldots ,{h_{10}} be in HP. If a1=h1=2  and  a10=h10=3{a_1} = {h_1} = 2\mathrm{\;and\;}{a_{10}} = {h_{10}} = 3 then a4h7{a_4}{h_7} is
View answer and explanation
Correct option: D
6
Option D (6) is correct because it is the product of the 4th term of the AP and the 7th term of the HP, calculated using the properties of AP and HP.
346 Medium
346 of 564
Application +4 | -1
If (am)n=amn\left(a^{m}\right)^{n}=a^{m^{n}}, then the value of ' m′\mathrm{m}^{\prime} in terms of ' n\mathrm{n} ' is
View answer and explanation
Correct option: C
n1/(n−1)n^{1 /(n-1)}
Option C (n1/(n−1)n^{1 /(n-1)}) is correct because solving the equation (am)n=amn(a^{m})^{n}=a^{m^{n}} for 'm' in terms of 'n' yields m=n1n−1m=n^{\frac{1}{n-1}}.
347 Medium
347 of 564
Application +4 | -1
Let b1,b2,……….b_1, b_2, \ldots \ldots \ldots .. , be a geometric sequence such that b1+b2=1b_1+b_2=1 and ∑k=1∞bk=2\sum\limits_{{k=1}}^{\infty} b_k=2. Given that b2<0b_2<0, then the value of b1b_1 is:
View answer and explanation
Correct option: C
2+22+\sqrt{2}
Option C (2+22+\sqrt{2}) is correct because it's derived by solving the equations formed by the given conditions of the geometric sequence and considering the constraint b2 < 0.
348 Medium
348 of 564
Application +4 | -1
If the sum of nn terms of an AP is 3n2−n3 n^2-n and its common difference is 6 , then its first term is
View answer and explanation
Correct option: A
2
Option A (2) is correct because by finding the general term of the AP using the sum of n terms, we can determine the first term.
349 Medium
349 of 564
Analyzing +4 | -1
If a,ba, b and cc are in HP, then correct statement is
View answer and explanation
Correct option: B
a2+c2>2b2a^2+c^2>2 b^2
Option B (a2+c2>2b2a^2+c^2>2 b^2) is correct because the Arithmetic Mean (AM) is always greater than the Harmonic Mean (HM), and this relationship can be used to derive the inequality.
350 Medium
350 of 564
Understanding +4 | -1
If the arithmetic mean and geometric mean of aa and bb are AA and GG respectively, then the value of A−GA-G will be
View answer and explanation
Correct option: C
[a−b2]2\left[\frac{\sqrt{a}-\sqrt{b}}{\sqrt{2}}\right]^2
Option C (a−b22\frac{{\sqrt{{a}} - \sqrt{{b}}}}{{\sqrt{{2}}}}^2) is correct because it is the simplified form of the difference between the arithmetic mean and geometric mean of a and b.
351 Medium
351 of 564
Application +4 | -1
If a,ba, b and cc are in AP\mathrm{AP}, then (a+2b−c)(2b+c−a)(a+2 b-c)(2 b+c-a) (c+a−b)(c+a-b) equals
View answer and explanation
Correct option: D
4abc4 a b c
Option D (4abc4abc) is correct because the expression simplifies to 4abc4abc when 2b=a+c2b = a + c is substituted, given that a, b, and c are in AP.
352 Medium
352 of 564
Application +4 | -1
Let pnp_n denote the probability of getting nn heads when a fair coin is tossed mm times. If P4,P5,P6P_4, P_5, P_6 are in Arithmetic Progression, then the sum of all possible values of mm is
View answer and explanation
Correct option: B
21
Option B (21) is correct because by setting up the arithmetic progression condition for the probabilities and solving for m, the sum of all possible values of m is found to be 21.
353 Medium
353 of 564
Application +4 | -1
The interior angles of a convex polygon form an arithmetic progression with a common difference of 4∘4^{\circ}. Number of sides of the polygon if its largest interior angle is 172∘172^{\circ} are
View answer and explanation
Correct option: D
12
Option D (12) is correct because the number of sides of the polygon is found by solving the arithmetic progression formula and the sum of interior angles formula.
354 Medium
354 of 564
Application +4 | -1
The natural numbers are written as follows

12345678910............\begin{array}{cccccccccccccccccccc} &&&1 \\ &&2&&3& \\ &4&&5&&6 \\ 7&&8&&9&&{10} \\ \begin{gathered} . \\ . \\ . \\ \end{gathered} &&\begin{gathered} . \\ . \\ . \\ \end{gathered} &{}&\begin{gathered} . \\ . \\ . \\ \end{gathered} &{}&\begin{gathered} . \\ . \\ . \\ \end{gathered} \end{array}

The sum of the numbers in the nthn^{\text {th}} row is
View answer and explanation
Correct option: B
n2(n2+1)\frac{n}{2}\left(n^2+1\right)
Option B is correct because the sum of the numbers in the nth row is found by recognizing the arithmetic progression and applying the sum formula.
355 Medium
355 of 564
Analyzing +4 | -1
If a,a1,a2,a3,…,a2n,ba, a_1, a_2, a_3, \ldots, a_{2 n}, b are in AP and a,g1,g2,…,g2n,ba, g_1, g_2, \ldots, g_{2 n}, b are in GP and hh is the HM of aa and bb, then a1+a2ng1g2n+a2+a2n−1g2g2n−1+…+an+an+1gngn+1\frac{a_1+a_{2 n}}{g_1 g_{2 n}}+\frac{a_2+a_{2 n-1}}{g_2 g_{2 n-1}}+\ldots+\frac{a_n+a_{n+1}}{g_n g_{n+1}} is equal to
View answer and explanation
Correct option: D
2nh\frac{2 n}{h}
Option D (2nh\frac{2 n}{h}) is correct because after using the properties of AP and GP, and the definition of HM, the expression simplifies to 2nh\frac{2 n}{h}.
356 Medium
356 of 564
Application +4 | -1
1.2.3+2.3.4+3.4.5+……+n1.2 .3+2.3 .4+3.4 .5+\ldots \ldots+\mathrm{n} terms -
View answer and explanation
Correct option: C
n(n+1)(n+2)(n+3)4\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(n+3)}{4}
Option C (n(n+1)(n+2)(n+3)4\frac{{n(n+1)(n+2)(n+3)}}{{4}}) is correct because the sum of the series is found by using the formula for summation of consecutive terms.
357 Medium
357 of 564
Application +4 | -1
If the sum of mm terms of an A.P. is the same as the sum of its nn terms, then the sum of its (m+n)(m+n) terms is .
View answer and explanation
Correct option: D
0
Option D (0) is correct because when the sum of m terms equals the sum of n terms in an A.P., the sum of (m+n) terms is zero.
358 Medium
358 of 564
Application +4 | -1
If ppth, qqth and rrth terms of a G.P. are a,b,c;a, b, c; then (cb)p(ba)r(ac)q=\left(\frac{c}{b}\right)^p\left(\frac{b}{a}\right)^r\left(\frac{a}{c}\right)^q=
View answer and explanation
Correct option: B
1
Option B (1) is correct because when the given expression is simplified using the properties of exponents and the general term of a G.P., the result is 1.
359 Medium
359 of 564
Analyzing +4 | -1
In an increasing geometric series, the sum of the second and the sixth term is 252\frac{{25}}{2} and the product of the third and fifth term is 25. Then, the sum of 4th, 6th and 8th terms is equal to
View answer and explanation
Correct option: C
35
Option C (35) is correct because by setting up equations based on the given information about the geometric series, we can solve for the first term and common ratio, and then calculate the sum of the 4th, 6th, and 8th terms.
360 Medium
360 of 564
Application +4 | -1
If log⁡32,log⁡3(2x−5)\log _3 2, \log _3\left(2^x-5\right) and log⁡3(2x−7/2)\log _3\left(2^x-7 / 2\right) are in AP, then xx is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because it satisfies the condition for the given logarithmic terms to be in arithmetic progression, after checking for valid logarithmic arguments.
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