Sequences & Series Questions - JEE

If $a_1, a_2, \ldots, a_n$ are in H.P., then $\dfrac{a_1}{a_2 + a_3 + \cdots + a_n},\ \dfrac{a_2}{a_1 + a_3 + \cdots + a_n},\ \ldots,\ \dfrac{a_n}{a_1 + a_2 + \cdots + a_{n-1}}$ are in

Consider an A.P.: $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and $\displaystyle\sum_{i=1}^n a_i=\dfrac{525}{2}$, then $\displaystyle\sum_{i=1}^{17}a_i$ is equal to

Find the sum $11^2 - 1^2 + 12^2 - 2^2 + 13^2 - 3^2 + \cdots + 20^2 - 10^2$.

The sum of series $1 + \dfrac{4}{5} + \dfrac{7}{5^2} + \dfrac{10}{5^3} + \cdots$ to ∞ is

The sum of 20 terms of the series whose $r$th term is given by $T(n) = (-1)^n \dfrac{n^2 + n + 1}{n!}$ is

Let $S = \dfrac{4}{19} + \dfrac{44}{19^2} + \dfrac{444}{19^3} + \cdots$ up to $\infty$. Then $S$ is equal to

The sum of first 9 terms of the series $\dfrac{1^3}{1} + \dfrac{1^3+2^3}{1+3} + \dfrac{1^3+2^3+3^3}{1+3+5} + \cdots$ is

If $ab^2c^3$, $a^2b^3c^4$, $a^3b^4c^5$ are in A.P. ($a, b, c > 0$), then the minimum value of $a + b + c$ is

Find the sum $3 + 7 + 14 + 24 + 37 + \cdots$ up to 20 terms.

If $\displaystyle\sum_{r=1}^{n} T_r = \frac{n}{8}(n+1)(n+2)(n+3)$, then find $\displaystyle\sum_{r=1}^{n}\frac{1}{T_r}$.

The sum of the series $\dfrac{1}{2!} - \dfrac{1}{3!} + \dfrac{1}{4!} - \cdots$ upto infinity is

If $a_1, a_2, a_3, \ldots, a_{2n+1}$ are in A.P., then $\dfrac{a_{2n+1} - a_1}{a_{2n+1} + a_1} + \dfrac{a_{2n} - a_2}{a_{2n} + a_2} + \cdots + \dfrac{a_{n+2} - a_n}{a_{n+2} + a_n}$ is equal to

The number of points, having both co-ordinates as integers that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0), is

Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1 = 15, 27 - 2a_2 > 0$ and $a_i = 2a_{i-1} - a_{i-2}$ for $k = 3, 4, \ldots, 11$. If $\frac{a_1 + a_2 + \ldots + a_{11}}{11} = 90$, then the value of $\frac{a_2 + a_4 + \ldots + a_{11}}{11}$ is equal to

Let $S_n = \dfrac{1}{1^3} + \dfrac{1+2}{1^3+2^3} + \dfrac{1+2+3}{1^3+2^3+3^3} + \ldots\ldots (n \text{ terms})$, where $n = 1,2,3,4,\ldots\ldots$, then $S_n$ is always less than:

If $t_n$ denotes the $n$th term of the series $2 + 3 + 6 + 11 + 18 + \cdots$ then $t_{50}$ is

Write the first five terms of the following sequence and obtain the corresponding series.$a_1 = a_2 = 2,\ a_n = a_{n-1} - 1,\ n > 2$

Given $\dfrac{\left(\dfrac{a}{b}\right)}{\left(\dfrac{b-1}{b}\right)} = 4$, find $S = \dfrac{\left(\dfrac{a}{a+b}\right)}{\left(1 - \dfrac{1}{a+b}\right)} = \dfrac{a}{a+b-1}$.

If $S = \displaystyle\sum_{n=1}^{5} \frac{1}{n(n+1)(n+2)(n+3)} = \frac{k}{3}$, then $k$ equals:

For Problems 7–9: In a G.P., the sum of the first and last terms is 66, the product of the second and the last but one is 128, and the sum of the terms is 126.In any case, the difference of the least and greatest terms is

If $S_K = \dfrac{k^2 - 1}{1 - \dfrac{1}{k}} = k(k+1)$ for $k \neq 1$, and $S_1 = T_1 = 0$, find the sum $S = \displaystyle\sum_{k=2}^{\infty} \dfrac{k(k+1)}{2^{k-1}}$.

The sum of the first 20 terms common between the series $3 + 7 + 11 + 15 + \cdots$ and $1 + 6 + 11 + 16 + \cdots$, is

Find the sum $\displaystyle\sum_{n=2}^{\infty}\frac{3n^2+1}{(n^2-1)^3}$.

In a GP of positive terms, any term is equal to the sum of the next two terms. Which of the following is the common ratio?

The sum of the series $\dfrac{1^3}{1} + \dfrac{1^3+2^3}{1+3} + \dfrac{1^3+2^3+3^3}{1+3+5} + \cdots$ up to 9 terms is:

If n arithmetic means are inserted between 1 and 31 such that the 7th mean : the $(n-1)$th mean = 5 : 9, then find n.

Let k be natural number. Defined $ S_k $ as the sum of the infinite geometric series with first term $ (k^2 - 1) $ and common ratio $ \dfrac{1}{k} $, that is $ S_k = \dfrac{k^2-1}{k^0} + \dfrac{k^2-1}{k^1} + \dfrac{k^2-1}{k^2} + \cdots $. The value of $ \displaystyle\sum_{k=1}^{\infty} \dfrac{S_k}{2^{k-1}} $, is:

Find the sum up to $n = 15$ terms of the series \[1 + 6 + \frac{9(1^2+2^2+3^2)}{7} + \frac{12(1^2+2^2+3^2+4^2)}{9} + \frac{15(1^2+2^2+\cdots+5^2)}{11} + \cdots\]

Find the sum $(x+y) + (x^2 + xy + y^2) + (x^3 + x^2y + xy^2 + y^3) + \cdots$ $n$ terms.

The value of $(2)^{1/4} \cdot (4)^{1/8} \cdot (8)^{1/16} \cdots \infty$ is:

Let one AM $a$ and two GMs $g_1$ and $g_2$ be inserted between $b$ and $c$. Then $\dfrac{g_1^3 + g_2^3}{abc} =$

The ratio of their first term is

Let a12 =

6 3 3 6

Let the sum $\sum_{n=1}^{100} \frac{1}{n(n+1)(n+2)}$ written in the rational form be $\frac{p}{q}$ (where $p$ and $q$ are co-prime), then the value of $\lfloor \frac{5q}{4p} \rfloor$ is (where $\lfloor \cdot \rfloor$ is the greatest integer function)

If \(|a|

Let a_1, a_2, a_3, \ldots, a_{49} be in A.P. such that a_9 + a_{43} = 66.If \sum_{i=1}^{49} \frac{1}{a_i} is to be calculated, then m is equal to

If P(n) is a statement such that P(3) is true. Assuming P(k) \text{ is true } \Rightarrow P(k+1) \text{ is true for all } k \geq 3, then P(n) is true

Let n be the greatest integer for which 5p2 − 16, 2p, n − 2 are distinct consecutive terms of an AP, where p ∈ ℝ. If the common difference of the AP is $\frac{m}{n}$, where m, n ∈ ℕ and m, n are relatively prime, the value of m + n is

Let $6, (6+d), (6+2d), \ldots$ are in an A.G.P. $S_8 = (6+d)^8$ and $8 = (6+2d)^2$. Find $d$.

If $(1)(2020) + (2)(2019) + (3)(2018) + \cdots + (2020)(1) = 2020 \times 2021 \times k$, then the value of $\frac{1}{105}$ is equal to

A geometric progression has first term $\alpha$ and common ratio $\beta$. Given that $3 = 2\left(\dfrac{c}{a} - \dfrac{b}{a}\right)$ leads to $8\alpha^2 + 10\alpha - 3 = 0$, find the value of $3S$ where $S = \dfrac{\beta}{1-\alpha}$ is the sum of the infinite GP.

Given $100 and H.C.F. \((91, n) > 1$. The sum of all natural numbers $n$ such that $100 and H.C.F. \((91, n) > 1$ is:

We have $x = \displaystyle\sum_{n=0}^{\infty} a^n$, $y = \displaystyle\sum_{n=0}^{\infty} b^n$, $z = \displaystyle\sum_{n=0}^{\infty} c^n$, where $a, b, c$ are in AP and \(|a|

139. If $T_n$ denotes the $n^{\text{th}}$ term of an arithmetic progression such that $T_p=\dfrac{1}{q}$ and $T_q=\dfrac{1}{p}$, then which of the given option is necessarily a root to the equation $(p+2q-3r)x^2+(q+2r-3p)x+(r+2p-3q)=0$, given that $p+2q-3r\neq 0$?

The value of $\displaystyle\sum_{k=1}^{13} \dfrac{1}{\sin\!\left(\dfrac{\pi}{4} + \dfrac{(k-1)\pi}{6}\right)\sin\!\left(\dfrac{\pi}{4} + \dfrac{k\pi}{6}\right)}$ is equal to

In a cricket tournament 16 school teams participated. A sum of ₹8000 is to be awarded among themselves as prize money. If the last placed team is awarded ₹275 in prize money and the award increases by the same amount for successive finishing places, what amount will the first place team receive?

If $\displaystyle\sum_{r=1}^n T_r = \dfrac{n(n+1)(n+2)(n+3)}{12}$, where $T_r$ denotes the $r$-th term, then the value of $\displaystyle\lim_{n\to\infty}\sum_{r=1}^n \dfrac{1}{T_r}$ is

The AM of two given positive numbers is 2. If the larger number is increased by 1, the GM of the numbers becomes equal to the AM of the given numbers. Then, the HM of the given numbers is

[IIT JEE 2011] 30. If three successive terms of a G.P. with common ratio r(r > 1) are the lengths of the sides of a

Sequences & Series Questions (847)