Sequence of series practice · Easy

The number of solutions of the equation ${\log _4}\left( {x - 1} \right) = {\log _2}\left( {x - 3} \right),$ is

If $y + \frac{{{y^3}}}{3} + \frac{{{y^5}}}{5} + \ldots \infty = 2\left( {x + \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + \ldots \infty } \right),$ then

The sum to infinity of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\ldots$ is

If $x = 1 + 2 + \frac{4}{{2!}} + \frac{8}{{3!}} + \frac{{16}}{{4!}} + \ldots$, then ${x^{ - 1}}$ is equal to

If the $m$th term of HP be $n$ and $n$th term be $m$, then the $r$th term will be

If $a_1, a_2, a_3, \ldots \ldots$ is an A.P such that $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225 \text {, }$ then $a_1+a_2+a_3+\ldots \ldots+a_{23}+a_{24}$ is equal to

If $3^{2 \sin 2 \theta-1}, 14,3^{4-2 \sin 2 \theta}$ are first three terms of an A.P. for some $\theta$, then the sixth term of this A.P. is

The first two terms of a geometric progression add upto 12. The sum of the third and the fourth terms is 48. If terms of the geometric progression are alternately positive and negative, then the first term is

Three non-zero real numbers form an A.P. and the squares of these numbers taken in the same order form a G.P. Then, then number of all possible values of common ratios of the G.P. is

Let the positive numbers $a,\;b,\;c,\;d$ be in AP, then $abc,\;abd,\;acd,\;bcd$ are

The product ${2^{\frac{1}{4}}} \cdot {4^{\frac{1}{{16}}}} \cdot {8^{\frac{1}{{48}}}} \cdot {16^{\frac{1}{{128}}}} \cdot ....\,\mathrm{to }\,\,\infty$ is equal to

The sum of series $1 - 3 + 5 - 7 + 9 - 11 + \ldots$ to $n$ terms is

For any three positive real numbers $a, b$ and $c$, $9\left(25 a^2+b^2\right)+25\left(c^2-3 a c\right)=15 b(3 a+c)$. Then

If $\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)$ and $\log \left(\frac{a}{3 b}\right)$ are in A.P., where $a, b, c$ are in G.P., then $a, b, c$ are the lengths of sides of

Sum of the series $P=\frac{1}{2 \sqrt{1}+\sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\ldots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}}$ is

If the sum of the series $2,5,8,11, \ldots \mathrm{is}\;\mathrm{\;}60100,$ then $n,$ the number of terms, is

A person is to count 45000 currency notes. Let ${a_n}$ denotes the number of notes he counts in the $n$ th minute. If ${a_1} = {a_2} = ... = {10_{10}} = 150$ and ${a_{10}},{a_{11}}, \ldots$ are in AP with common difference -2, then the time taken by him to count all notes, is

Suppose $\theta$ and $\phi(\neq 0)$ are such that $\sec (\theta+\phi), \sec \theta$ and $\sec (\theta-\phi)$ are in A.P. If $\cos \theta=k \cos \left(\frac{\phi}{2}\right)$ for some $k$, then $k$ is equal to

The sum of the series $\frac{1}{2}{x^2} + \frac{2}{3}{x^3} + \frac{3}{4}{x^4} + \frac{4}{5}{x^5} + \ldots$ is

The sum of the series $\frac{4}{{1\;!}} + \frac{{11}}{{2\;!}} + \frac{{22}}{{3\;!}} + \frac{{37}}{{4\;!}} + \frac{{56}}{{5\;!}} + \ldots ,$ is

If $[x]$ denotes the greatest integer $\leq x$, then value of $S=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{2024}]$ is

If $a,b,c$ be in A.P., $b,c,d$ are in G.P., and $c,d,e$ are in H.P., then $a,c,e$ will be in

Let ${T_r}$ be the $r$th term of an AP for$r = 1,\;2,\;3$,… If for some positive integers $m,\;n,\;\mathrm{we\;hav}e\;{T_m} = \frac{1}{n}\mathrm{\;and}\;{T_n} = \frac{1}{m},\;\mathrm{then}\;{T_{mn}}$ equals

If $f: R \rightarrow R$ satisfies $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=7$, then $\sum\limits_{r=1}^n f(r)$, is

If $m^{\text {th }}$ terms of the series $63+65+67+69+\cdots$ and $3+10+17+24+\cdots$ are equal, then $m=$

If ${\log _5}({\log _5}\left( {{{\log }_2}x} \right)) = 0,$ then the values of $x$ is

If $\sum\limits_{{n=1}}^n n, \frac{\sqrt{10}}{3} \cdot \sum\limits_{{n=1}}^n n^2$ and $\sum\limits_{{n=1}}^n n^3$ are in GP, then the value of $n$ is

If $\log _3 2, \log _3\left(2^x-5\right)$ and $\log _3\left(2^x-\frac{7}{2}\right)$ are in $\mathrm{AP}, x$ is equal to

If $a, b$ and $c$ are in GP, then $a^{1 / 2}, b^{1 / 2}, c^{1 / 2}$ are in

If $\left\{ {{a_i}} \right\}_{i = 1}^n,$ where $n$ is an even integer, is an arithmetic progression with common difference 1, and $\sum\limits_{i = 1}^n {{a_i}} = 192,\,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}}} = 120,$ then $n$ is equal to: