Sequence of series practice · Mixed difficulty

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

Two sequences $< {a_n} >$ and $< {b_n} >$ are defined by ${a_n} = \log \left( {\frac{{{5^{n + 1}}}}{{{3^{n - 1}}}}} \right),{b_n} = {\left\{ {\log \left( {\frac{5}{3}} \right)} \right\}^n},$ then

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 least positive integer $\mathrm{n}$ such that $1-\frac{2}{3}-\frac{2}{3^{2}}-\ldots .-\frac{2}{3^{\mathrm{n}-1}}<\frac{1}{100}$, is:

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

The coefficient of ${x^n}$ in the series $1 + \frac{{a + bx}}{{1!}} + \frac{{{{\left( {a + bx} \right)}^2}}}{{2!}} + \frac{{{{\left( {a + bx} \right)}^3}}}{{3!}} = \ldots \infty$ is

If ${a_1},\;\;{a_2},\;\;{a_3}, \ldots ,\;{a_{24}}$ are in arithmetic progression and ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225,$ then
${a_1} + {a_2} + {a_3} + ... + {a_{23}} + {a_{24}}$ is equal to

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

If ${\log _4}\left( {3{x^2} + 11x} \right) > 1,$ then $x$ lies in the interval

${\log _e}\frac{{1 + 3x}}{{1 - 2x}}$ is equal to

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 $a, b, c, d, e$ and $f$ are in AP, then $e=c$ is equal to

If ${\log _5}({\log _5}\left( {{{\log }_2}x} \right)) = 0,$ then the values of $x$ 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 the sum to infinity of the series $1+4 x+7 x^2+10 x^3+\ldots \ldots \ldots$ is $\frac{35}{16}$, where $|x|<1$ then ${x}$ is equal to

Sum of the scries $9+99+999+\ldots$ upto $n$ terms is

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:

Let $A_1, A_2$ be two arithmetic means, $G_1$, $G_2$ be two geometric means, and $H_1, H_2$ be two harmonic means between two positive numbers $a$ and $b$, then values $\frac{G_1 G_2}{H_1 H_2} \cdot \frac{H_1+H_2}{A_1+A_2}$ is

The $(m+n)$ th and $(m-n)$ th terms of a GP are $p$ and $q$, respectively. Then, the $m$ th term of the GP is

If the $2^{\text {nd }}, 5^{\text {th }}$ and $9^{\text {th }}$ terms of a non-constant A.P. are in G.P., then the common ratio of this G.P., is

If the sum of $m$ consecutive odd integers is $m^4$, then the first integer is