Sequence of series practice · Medium

The sum of all possible products of the first $n$ natural mumbers taken two by two, is

Find the range of values of $x$ that satisfy the following system of inequations.

$-17 \leq 3 x+10 \leq-2 ;$

$-22 \leq 5 x+13 \leq 3$ and

$-19 \leq 2 x-9 \leq-3$

The $n^{\text {th}}$ term of series $\frac{1}{1}+\frac{1+2}{2}+\frac{1+2+3}{3}+\ldots$ will be

Let ${a_1},{a_2},.....{a_{30}}$ be an AP, $S = \sum\limits_{i = 1}^{30} {{a_i}}$ and $T = \sum\limits_{i = 1}^{15} {{a_{(2i - 1)}}}$ If ${a_5} = 27\mathrm{ \,and}\,\,\mathrm{ }S - 2T = 75,\mathrm{then}\,\mathrm{ }\,\,{a_{10}}$ is equal to

$\begin{array}{cccccccccccccccccccc} {{{\left( {666 \ldots 6} \right)}^2}}\\ {n\;\mathrm{digits}} \end{array} + \begin{array}{cccccccccccccccccccc} {\left( {888 \ldots 8} \right)}\\ {n\;\mathrm{digits}} \end{array}$ is equal to

If $\frac{1}{2}, \frac{1}{2^2}, \frac{1}{2^3}, \frac{1}{2^4}, \ldots$ are in $G P$, then $2,2^2, 2^3, \ldots$ are in

If $b$ is the first term of an infinite $\mathrm{G} . \mathrm{P}$ whose sum is five, then $b$ lies in the interval.

Let $A_{n}=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^{2}+\left(\frac{3}{4}\right)^{3}-\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^{n}$ 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

$96^{\text {th }}$ term of the sequence $2,7,13,20,28, \ldots$. Is

$\sum \limits_{n = 1}^\infty \frac{{2{n^2} + n + 1}}{{n!}}$ is equal to

If sum of $n$ terms of an AP is $2n + 3{n^2}$ then $r$th term is

If sum of first 20 terms of an A.P. is equal
to sum of first 30 terms of the A.P. then sum of the first 50
terms of the A.P. is

If a function $f(x)$ satisfies $f(x+y)=f(x) \cdot f(y) ; \forall \, x, y \in \mathrm{N}$ and $f(1)=3$ and $\sum\limits_{{x=1}}^n f(x)=120$, then $n=$

The sum of the integers from 1 to 100 which are not divisible by 3 or 5 is

If $\frac{{{e^{5x}} + {e^x}}}{{{e^{3x}}}}$ is expand in a series of ascending powers of $x$ and $n$ is an odd natural number, then the coefficient of ${x^n},$ is

Let $a$ and $b$ be two distinct real numbers. Let $11^{\text {th }}$ term of a GP, whose first term is $a$ and third term is $b$, is equal to $p^{\text {th }}$ term of another GP. Whose first term is $a$ and fifth term is $b$. Then $p$ is equal to

The value of ${3^{\frac{4}{{{{\log }_4}9}}}} + {27^{\frac{1}{{{{\log }_{36}}9}}}} + {81^{\frac{1}{{{{\log }_5}3}}}},\;\mathrm{is}$

Sum of the series $\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\ldots \ldots$ upto $n$ terms is equal to

If $x_1, x_2, x_3$ as well as $y_1, y_2, y_3$ are in GP with same common ratio, then the points $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$

If the lengths of sides of a right triangle are in AP, then the sines of the acute angle are

If $2 \times {2^2} + 3 \times {2^3} + 4 \times {2^4} + \ldots + n \times {2^n} = {2^{n + 10}},$ then $n =$

If $a, b$, $c$ are in G.P., $x$ and $y$ be the A.M.'s between $a, b$ and $b, c$ respectively, then $\left(\frac{a}{x}+\frac{c}{y}\right)\left(\frac{b}{x}+\frac{b}{y}\right)$ is equal to

Let $s_{1}, s_{2}, s_{3} \ldots, s_{10}$ respectively be the sum of 12 terms of 10 A.Ps whose first terms are $1,2,3, \ldots ., 10$ and the common differences are $1,3,5, \ldots, 19$ respectively. Then $\sum\limits_{{i=1}}^{10} s_{i}$ is equal to

The length of a side of a square is $a$ metre. A second square is formed by joining the mid-points of these square. Then, a third square is formed by joining the mid-points of the second square and so on. Then, the sum of the area of the squares which carried upto infinity, is

If $S_r$ denotes the sum of the first $r$ terms of an $A P$, then $\frac{S_{3 r}-S_{r-1}}{S_{2 r}-S_{2 r-1}}$ is equal to

The sum of the first $n$ terms of the series $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\ldots \ldots \ldots \ldots$ is .

$11^{3}+12^{3}+\ldots +20^{3}$

If $-\pi / 2 < x < \pi / 2$, and the sum to infinite terms of the series $\cos x+\frac{2}{3} \cos x \sin ^2 x+\frac{4}{9} \cos x \sin ^4 x+\ldots$ is finite then

If $(1.05)^{50}=11.658$, then $\sum_{n=1}^{49}(1.05)^n$ equals

If $x+y+z=1$ and $x, y$ and $z$ are positive numbers such that $(1-x)(1-y)(1-z) \geq k x y z$, then $k$ is equal to