Sequences & Series Questions (847)

The roots of the equation $3x^2-px+q=0$ are the $10^{\text{th}}$ and $11^{\text{th}}$ terms of an AP with common difference $\dfrac{3}{2}$ and $S_{11}=88$. Then $q-2p$ equals
Let the first term of an AP be 3. If the sum of its first 4 terms is $\dfrac{1}{5}$ of the sum of the next 4 terms, then $S_{20}$ equals
Let $a_0=0$, $a_1=\dfrac{1}{2}$, and $2a_{n+2}=5a_{n+1}-3a_n$ for $n\geq0$. Then $\displaystyle\sum_{k=1}^{100}a_k$ equals
Let $a,ar,ar^2,\ldots$ be an infinite G.P. If $\sum_{n=0}^{\infty}ar^n=57$ and $\sum_{n=0}^{\infty}a^3r^{3n}=9747$, then $a+18r$ is equal to
If the sum of the series $\dfrac{1}{1\cdot(1+d)}+\dfrac{1}{(1+d)(1+2d)}+\cdots+\dfrac{1}{(1+9d)(1+10d)}$ is equal to 5, then 50d is equal to:
320. The product of \(n\) positive numbers is unity. Then their sum is
Suppose that \(F(n+1) = \dfrac{2F(n)+1}{2}\) for \(n = 1, 2, 3, \ldots\) and \(F(1) = 2\). Then \(F(101)\) equals
29. If \(a^2 + b^2\), \(ab + bc\), and \(b^2 + c^2\) are in G.P., then \(a, b, c\) are in
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is:
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
Let$A = {1$, 6, 11, 16,$\ldots} and$B = {9$, 16, 23, 30,$$\ldots} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A$$\cup B) is$
4 1 1 1$\pi If + + +$$\ldots..$$\infty =, 4 4 4 1 2 3 90 1 1 1 + + +$$\ldots..$$\infty =$$\alpha, 4 4 4 1 3 5 1 1 1 + + +$$\ldots.$$\infty =$$\beta, 4 4 4 2 4 6 then$$\alpha$$\beta is equal to$
Let \(A_n = \left(\dfrac{3}{4}\right) - \left(\dfrac{3}{4}\right)^2 + \left(\dfrac{3}{4}\right)^3 - \cdots + (-1)^{n-1}\left(\dfrac{3}{4}\right)^n\) and \(B_n = 1 - A_n\). Then the least odd natural number \(p\), so that \(B_n > A_n\) for all \(n \geq p\), is
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
Let $S_n = \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdots$ to $n$ terms. If the sum of the first 6 terms of an AP with first term $-p$ and common difference $p$ equals $\sqrt{2026\cdot S_{2025}}$, and an denotes the $n^{\text{th}}$ term of this AP, then $|A_{20}-A_{15}|$ equals
The sum $\displaystyle\sum_{n=1}^{\infty}\frac{9n^2}{(3n)!}$ is equal to
Find the number of common terms in the following sequences: (ii) 1, 5, 9, … to 100 terms and 4, 7, 10, … to 60 terms.
The value of $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$ to $\infty$ is
If $1+\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}+\dfrac{5-2\sqrt{6}}{18}+\dfrac{9\sqrt{3}-11\sqrt{2}}{36\sqrt{3}}+\dfrac{49-20\sqrt{6}}{180}+\cdots$ upto $\infty=2+\left(\sqrt{\dfrac{b}{a}}+1\right)\log_e\!\left(\dfrac{a}{b}\right)$, where $a$ and $b$ are integers with $\gcd(a,b)=1$, then $11a+18b$ is equal to
Let one AM $a$ and two GMs $g_1$ and $g_2$ be inserted between $b$ and $c$. Then $\dfrac{g_1^3 + g_2^3}{abc} =$
Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference $l$ are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1a_2a_3a_4+l^4=361$, then the largest term of the A.P. is equal to
$\dfrac{2^3-1^3}{1\times7}+\dfrac{4^3-3^3+2^3-1^3}{2\times11}+\dfrac{6^3-5^3+\cdots+1^3}{3\times15}+\cdots+\dfrac{30^3-29^3+\cdots+1^3}{15\times63}$ is equal to
If $\cot^{-1}\left(\dfrac{n^2-10n+21.6}{\pi}\right)>\dfrac{\pi}{6}$, then the number of positive integers $n$ satisfying this is
Let $x_1, x_2, \ldots, x_{100}$ be an A.P. with $x_1=1$ and $x_{100}=199$. If $y_i=i(x_i+1)$; $i=1,2,\ldots,100$, then mean of $y_1, y_2, \ldots, y_{100}$ is
Let $\{b_n\}_{n=0}^\infty$ be a sequence of positive real numbers such that $b_0=1$ and $b_n=2+b_{n-1}-2\sqrt{1+b_{n-1}}$. Then $\displaystyle\sum_{n=0}^\infty\frac{b_n}{2^n(n+1)}$ equals:
Let $\{b_n\}_{n=0}^\infty$ be a sequence of positive real numbers such that $b_0=1$ and $b_n=2+b_{n-1}-2\sqrt{1+b_{n-1}}$. Then $\displaystyle\sum_{n=0}^\infty\frac{b_n}{2^n(n+1)}$ equals:
if $p^q$, $q^r$ and $r^p$ terms of an H.P. be respectively $x$, $y$, $z$, then $(p-q)xy + (q-r)yz + (r-p)xz=$
If $\frac{-3+12}{1+3+...+(2l-1)} = \frac{\text{some}}{(2l-1)\text{ some}}$, then $n$ is equal to
The sum to 10 terms of the series $\dfrac{1}{1 + 1^2 + 1^4} + \dfrac{2}{1 + 2^2 + 2^4} + \dfrac{3}{1 + 3^2 + 3^4} + \ldots$ is:
If the $(m+1)^{th}$, $(n+1)^{th}$ and $(r+1)^{th}$ terms of an A.P. are in G.P. and $m,n,r$ are in H.P., then the ratio of the common difference to the first term in the A.P. is equal to:
The third term of a G.P. is 2. Then the product of the first five terms is
The geometric mean of 6 observations was calculated as 13. It was later observed that one of the observations was recorded as 28 instead of 36. The correct geometric mean is
If $a_n = \dfrac{-2}{4n^2 - 16n + 15}$, then $a_1 + a_2 + \ldots + a_{25}$ is equal to:
If 1+3+5+....upto n terms __ 20 and 4+7+10+...uptonterms 7log,,x i i i n= log, x+tlog,, x? +log,, x* +log,, x® +.....+00, then x is equal to
$\frac{n}{1 \cdot 2 \cdot 3} + \frac{n-1}{2 \cdot 3 \cdot 4} + \frac{n-2}{3 \cdot 4 \cdot 5} + \ldots$ upto $n$ terms is equal to:
If $a, b, c, d, e, f$ are in arithmetic progression. Then $e - c$ is equal to
$\sum_{r=1}^{99} r(r^2 + r + 1)$ is equal to:
$\left(\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3^2}+\dfrac{1}{3}\times\dfrac{4}{7}+\dfrac{4^2}{7^2}\right)+\left(\dfrac{1}{3^3}+\dfrac{1}{3^2}\times\dfrac{4}{7}+\dfrac{1}{3}\times\dfrac{4^2}{7^2}+\dfrac{4^3}{7^3}\right)+\cdots$ upto infinite terms, is equal to
If a, b,c are in AP, then (a - c)? equals
The sum of values of $n$ for which $S_n$ vanishes is:
Let $y = f(x)$ represent a parabola with focus $\left(-\dfrac{1}{2}, 0\right)$ and directrix $y = -\dfrac{1}{2}$. Then $S = \left\{x \in \mathbb{R} : \tan^{-1}\left(\sqrt{f(x)}\right) + \sin^{-1}\left(\sqrt{f(x)+1}\right) = \dfrac{\pi}{2}\right\}$:
The value of $\frac{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4})...(2n-1)^4 + \frac{1}{4})}{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4})...(2n)^4 + \frac{1}{4})}$ is equal to:
Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 1? + 2-2? + 37+ 2-47 +57 + 2-6" +....... If B-2A = 100A, then A is equal to : [JEE (Main) 2018] (1) 248 (2) 464 (3) 496 (4) 232
$\sum_{r=0}^{30} \frac{1}{Q(r)}$ is equal to:
The value of $\sum_{r=5}^{n} \frac{1}{t_r}$ is equal to:
Let $a_1, a_2, a_3, \ldots$ be an A.P. If $a_7 = 3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n! - 4a_{n(n+2)}$ is equal to:
If three distinct numbers a,b,c are in G.P. and the equations ax?+2bx+c=0 and dx? + 2ex + f = 0 have acommon root, then which one of the following statements is correct? [JEE (Main) 2019] f (1) d,e, f are in AP. are in GP. ass, c f are in A.P. (4) d,e, f are in G.P. @ 2S, a
32. 4n kk+1) Let the first term of a series be T, = 6 and its r term T, = 3T,_; + 6,,7 = 2, 3, _ n. Ifthe sum of [JEE (Advanced) 2013] Let S,=)°(-1) ? &’ . Then S;, can take value(s) ka the first n terms of this series is (0° -12n +39)(4.6" —5.3" +1). Then n is equal to
Let $\displaystyle\sum_{k=1}^n a_k=\alpha n^2+\beta n$. If $a_{10}=59$ and $a_6=7a_1$, then $\alpha+\beta$ is equal to
The sum to the infinite terms of the series $\frac{1}{2 \cdot 7} + \frac{1}{7 \cdot 12} + \frac{1}{12 \cdot 17} + ...$is