Sequences & Series Questions - JEE

Let $\alpha,\beta$ be roots of $x^2-10x+2=0$. Value of $\dfrac{\alpha^{2028}+\beta^{2028}+8\alpha^{2022}+8\beta^{2022}}{\alpha^{2025}+\beta^{2025}}$ is

If sum of the series $1+\dfrac{\sqrt5-\sqrt3}{2\sqrt5}+\dfrac{8-2\sqrt{15}}{30}+\dfrac{14\sqrt5-18\sqrt3}{60\sqrt5}+\cdots=2+\dfrac{a+\sqrt{15}}{b}\log_b\!\left(\dfrac{a}{c}\right)$; $a,b,c\in\mathbb{N}$, $\gcd(a,b,c)=1$, then $2(a^2+b^2+c^2)$ is

Let $a_1, a_2, a_3, \ldots$ be in harmonic progression with $a_1 = 5$ and $a_{20} = 25$. If $n$ is the least positive integer for which $a_n < 0$, then the value of $4n - 100$ is

Let $f_n(x) + f_n(y) = \frac{x^n + y^n}{x^n y^n}$ for all $x, y \in \mathbb{R} - \{0\}$ where $n \in \mathbb{N}$.Let $g(x) = \max\left\{f_2(x), f_3(x)\right\}$ for all $x \in \mathbb{R} - \{0\}$.The minimum value of $\sum_{k=1}^{\infty} f_{2k}(\csc \theta) + \sum_{k=1}^{\infty} f_{2k}(\sec \theta)$, where $\theta \neq \frac{k\pi}{2}; k \in \mathbb{I}$ is:

Let $s_1,s_2,\ldots,s_{10}$ be the sums of 12-term APs with first terms $1,2,\ldots,10$ and common differences $1,3,5,\ldots,19$. Then $\sum_{i=1}^{10}s_i$ is equal to

Consider an A.P. of positive integers, whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800$. Then its $11^{\text{th}}$ term is:

The value of $\dfrac{1\times2^2+2\times3^2+\cdots+100\times101^2}{1^2\times2+2^2\times3+\cdots+100^2\times101}$ is

Let an AP have $2k$ terms. If the sum of odd-position terms (1st, 3rd, 5th, ...) is 40, the sum of even-position terms is 55, and the last term minus the first term is 27, then $k$ equals

39. Let $a = 1\,1\,1\,\ldots1$ (55 digits), $b = 1 + 10 + 10^2 + \cdots + 10^4$, $c = 1 + 10^5 + 10^{10} + 10^{15} + \cdots + 10^{50}$, then

Let $n \in N, n > 25$. Let $A, G, H$ denote the arithmetic mean, geometric mean, and harmonic mean of 25 and $n$. The least value of $n$ for which $A, G, H \in \{25, 26, \ldots, n\}$ is

The sum of the series $2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \cdots + 62 \cdot {}^{20}C_{20}$ is equal to:

An arithmetic progression is written in the following way: Row 1: 2; Row 2: 5,8; Row 3: 11,14,17; Row 4: 20,23,26,29; $\ldots$ The sum of all the terms of the 10th row is

The number of common terms in the progressions $4,9,14,19,\ldots$ up to $25^{\text{th}}$ term and $3,6,9,12,\ldots$ up to $37^{\text{th}}$ term is:

How many three – term harmonic progressions a, b, c of strictly increasing positive integers in which $a = 20$ and $b$ divides $c$ exist?

166. If $a+c,\ a+b,\ b+c$ are in G.P. and $a, c, b$ are in H.P. where $a, b, c > 0$, then the value of $\dfrac{a+b}{c}$ is:

Let $a_{1},a_{2},a_{3},\dots$ be a G.P.\ of increasing positive terms. If $a_{1}a_{5}=28$ and $a_{2}+a_{4}=29$, then $a_{6}$ is equal to:

The value of $\displaystyle\sum_{r=1}^{\infty}\frac{8r}{4r^4+1}$ is

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10}=390$ and the ratio of the tenth and the fifth terms is $15:7$, then $S_{15}-S_5$ is equal to:

If three successive terms of a G.P. with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to:

41. The sum of 20 terms of a series of which every even term is 2 times the term before it, and every odd term is 3 times the term before it, the first term being unity is

If $7=5+\dfrac{5+\alpha}{7}+\dfrac{5+2\alpha}{7^{2}}+\dfrac{5+3\alpha}{7^{3}}+\dotsb\infty$, then the value of $\alpha$ is:

If $7 = 5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2\alpha)+\cdots$ to infinity, then $\alpha$ equals

Let $T_{r}$ be the $r^{\text{th}}$ term of an A.P. If for some $m$, $T_{m}=\dfrac{1}{25}$ and $T_{25}=\dfrac{1}{20}$, and $\displaystyle\sum_{r=1}^{25}T_{r}=\dfrac{13}{20}$, then $5m\,\displaystyle\sum_{r=m}^{2m}T_{r}$ is equal to:

In a non constant arithmetic progression having odd number of terms, having positive integral common difference, the ratio of the sum of the 1st, 3rd, 5th, 7th, ... terms to the sum of remaining terms is 13 : 12, then the number of terms in the arithmetic progression, is:

Let the first three terms 2, $p$ and $q$, with $q\neq2$, of a G.P. be respectively the 7th, 8th and 13th terms of an A.P. If the 5th term of the G.P. is the $n$th term of the A.P., then $n$ is equal to:

Let $a_{1},a_{2},\dots,a_{2024}$ be an Arithmetic Progression such that $a_{1}+(a_{5}+a_{10}+a_{15}+\dots+a_{2020})+a_{2024}=2233$. Then $a_{1}+a_{2}+\dots+a_{2024}$ is equal to \rule{2cm}{0.4pt}.

If $a$, $b$, $c$ are in AP and $(a+2b-c)(2b+c-a)(c+a-b)=\lambda abc$, then the value of $\lambda$ is

Let $x,y,z$ be three natural numbers such that $x+y+z=10$. The maximum possible value of $xyz+xy+yz+zx$ is

If $a$, $b$, $c$ are positive integers forming an increasing GP, $b-a$ is a perfect cube, and $\log_6 a + \log_6 b + \log_6 c = 6$, then $a+b+c$ is equal to

Given sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\dfrac{27}{19}$. Find the common ratio of the G.P.

$\dfrac{2^3-1^3}{1\times7}+\dfrac{4^3-3^3+2^3-1^3}{2\times11}+\dfrac{6^3-5^3+\cdots+1^3}{3\times15}+\cdots+\dfrac{30^3-29^3+\cdots+1^3}{15\times63}$ is equal to

If sum of the series $1+\dfrac{\sqrt5-\sqrt3}{2\sqrt5}+\dfrac{8-2\sqrt{15}}{30}+\dfrac{14\sqrt5-18\sqrt3}{60\sqrt5}+\cdots=2+\dfrac{a+\sqrt{15}}{b}\log_b\!\left(\dfrac{a}{c}\right)$; $a,b,c\in\mathbb{N}$, $\gcd(a,b,c)=1$, then $2(a^2+b^2+c^2)$ is

Given that AD, BE, CF are the altitudes of triangle ABC and are in HP, then which of the following is true?

If $a_1, a_2, a_3, \ldots, a_{4001}$ are terms of an AP such that $\frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_{4000} a_{4001}} = 10$ and $a_2 + a_{4000} = 50$, then $|a_1 - a_{4001}|$ is equal to

The number of terms in an AP is even; the sum of the odd terms in it is 24 and that the even terms is 30. If the last term exceeds the first term by $10\dfrac{1}{2}$, then the number of terms in the AP is

Let $\dfrac{1}{x_1}, \dfrac{1}{x_2}, \ldots, \dfrac{1}{x_n}$ ($x_i \neq 0$ for $i = 1, 2, \ldots, n$) be in AP such that $x_1 = 4$ and $x_{21} = 20$. If n is the least positive integer for which $x_n > 50$, then $\displaystyle\sum_{i=1}^{n}\left(\dfrac{1}{x_i}\right)$ is equal to

Find the sum of the series $1 + \dfrac{3}{2} + \dfrac{7}{4} + \dfrac{15}{8} + \dfrac{31}{16} + \cdots$ up to 20 terms.

Sequence $\{a_n\}$ with $a_1=a_2=1$, $a_3=3$, $a_n=(n+3)a_{n-1}-na_{n-2}-(n-2)a_{n-3}$ for $n\ge 4$. Then $\dfrac{a_{2023}-a_{2020}}{2021!}=$

A sequence of positive terms A_1, A_2, A_3, \ldots, A_n satisfies the relation A_{n+1} = \frac{3(1 + A_n)}{3 + A_n}. Find the least integral value of A_1 for which the sequence is decreasing.

Given that $a_1, a_2, a_3, \ldots, a_n$ are in HP and $E = a_1 a_2 + a_2 a_3 + \cdots + a_{n-1} a_n$. If $d$ is the common difference of the corresponding AP, then $E$ equals:

Let $a_1, a_2, \ldots, a_{10}$ be a G.P. If $\dfrac{a_3}{a_1} = 25$, then $\dfrac{a_9}{a_5}$ equals:

In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49, and the sum of the first and the third term is 35. Then the first term of this geometric progression is:

The value of \ \left(\frac{8}{5}\right)^2 + \left(\frac{12}{5}\right)^2 + \left(\frac{16}{5}\right)^2 + \left(\frac{20}{5}\right)^2 + \cdots\ up to 10 terms is $\frac{16}{5}m$. Find $m$.

$e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots}$ equal to

Find the number of common terms to the two sequences 17, 21, 25, ..., 417 and 16, 21, 26, ..., 466.

Sum of the series $1^2\cdot2^2+2^2\cdot3^2+\cdots+n^2\cdot(n+1)^2$ is

If 9 harmonic means be inserted between 2 and 3, then the value of $\frac{A - 6H}{5}$ (where A is any of the AM's and H is the corresponding HM), is

Find the sum to n terms of the series $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \cdots$

The sum of the first n terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + 2 \cdot 6^2 + \cdots$ is $\dfrac{n(n+1)^2}{2}$ when n is even. When n is odd the sum is

If sum of the series $1+\dfrac{\sqrt5-\sqrt3}{2\sqrt5}+\dfrac{8-2\sqrt{15}}{30}+\dfrac{14\sqrt5-18\sqrt3}{60\sqrt5}+\cdots=2+\dfrac{a+\sqrt{15}}{b}\log_b\!\left(\dfrac{a}{c}\right)$; $a,b,c\in\mathbb{N}$, $\gcd(a,b,c)=1$, then $2(a^2+b^2+c^2)$ is

Sequences & Series Questions (847)