Sequence of series practice · Mixed difficulty

If for every $n \in \mathbf{N}$, sum to $n$ terms of an A.P. is $5 n^2+7 n$ then its 10th term is

Consider the sequence $1,2,2,3,3,3, \ldots$, where $n$ occurs $n$ times. The number that occurs as 2011 th terms is

If the sum of the first ten terms of the series ${\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + {\left( {4\frac{4}{5}} \right)^2} + ...,$ is $\frac{{16}}{5}m$, then $m$ is equal to

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

If $m$ is the A.M. of two distinct real numbers $l$ and $n$ $(l, n>1)$ and $G_1 G_2$ and $G_3$ are three geometric means between $l$ and $n$, then $G_1{ }^4+2 G_2{ }^4+G_3{ }^4$ equals

If three numbers are in G.P., then the numbers obtained by adding the middle number to each of these numbers are in

If 1, ${\log _3}\sqrt {\left( {{3^{1 - x}} + 2} \right)} ,{\log _3}\left( {{{4.3}^x} - 1} \right)$ are in AP, then $x\;$ equals

Let $a_n$ be the $n$th term of an AP, if $\sum\limits_{r = 1}^{100} {{a_{2r}}} = \alpha \,\,{\text{and}}\,\,\,\sum\limits_{r = 1}^{100} {{a_{2r - 1}}} = \beta ,$ then the common difference of the AP is

An infinite G.P. has first term ' $x$ ' and sum '5', then $x$ belongs to

If $p, x_1, x_2 \ldots x_i, \ldots$ and $q, y_1, y_2 \ldots, y_i, \ldots$ are in A.P., with common difference $a$ and $b$ respectively, then the centre of mean position of the points $A_i\left(x_i, y_i\right)$, $i=1,2 \ldots n$ lies on the line

$\frac{1}{{n!}} + \frac{1}{{2!\left( {n - 2} \right)!}} + \frac{1}{{4!\left( {n - 4} \right)!}} + ...\infty$ is

If $a^2, b^2, c^2$ are in A.P. then $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in

If the first two terms of a H.P. are $2 / 5$ and $12 / 23$ respectively. Then, the largest term is

If $|\alpha|<1,|\beta|<1$
$1-\alpha+\alpha^2-\alpha^3+\ldots$. to $\infty=s_1$

$1-\beta+\beta^2-\beta^3+\ldots$ to $\infty=s_2$, then
$1-\alpha \beta+\alpha^2 \beta^2-\alpha^3 \beta^3+\ldots$ to $\infty$ equals

If ${\log _6}\left\{ {{{\log }_4}\left( {\sqrt {x + 4} + \sqrt x } \right)} \right\} = 0,$ then $x =$

Fifth term of an GP is 2, then the product of its 9 term is

The $n\mathrm{th}$ term of the sequence $4,14,30,52,80,114, \ldots ,$ is

A student read common difference of an AP as -3 instead of 3 and obtained the sum of first 10 terms as -30. Then, the actual sum of first 10 terms is equal to

Sum of the series $\sum\limits_{{r=1}}^n r \log \frac{r+1}{r}$ is

The value of $1 - {\log _e}2 + \frac{{{{\left( {{{\log }_e}2} \right)}^2}}}{{2\;!}} - \frac{{{{\left( {{{\log }_e}2} \right)}^3}}}{{3\;!}} + \ldots ,$ is

The determinant $\Delta = \left| {\begin{array}{cccccccccccccccccccc} a&b&{a\alpha + b}\\ b&c&{b\alpha + c}\\ {a\alpha + b}&{b\alpha + c}&0 \end{array}} \right|$ is equal to zero, if

Along a road lies an odd number of stones placed at intervals of $10 \mathrm{~m}$. These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried out the job starting with the stone in the middle, carrying stones in succession, thereby covering a distance of $4.8 \mathrm{~km}$. Then, the number of stones is

Let $S_n$ denote the sum of first $n$ terms of an A.P.If $S_{2 n}=3 S_n$ then the ratio $S_{3 n} / S_n$ is equal to

If ${a_1},{a_2},{a_3}, \ldots ,{a_n}$ be an AP of non-zero terms, then $\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ... + \frac{1}{{{a_{n - 1}}{a_n}}}$ is equal to

If the fourth power of the common difference of an A.P. with integer entries is added to the product of any four consecutive terms of it, then the resulting sum is

Let $\alpha ,\beta$ are the roots of $f\left( x \right) = a{x^2} + bx + c,\;a \ne 0$ and $\Delta = {b^2} - 4ac.\;$ If $\alpha + \beta ,\;{\alpha ^2} + {\beta ^2}\;\mathrm{and}\;{\alpha ^3} + {\beta ^3}$ are in GP, then

$1 + \frac{3}{2} + \frac{5}{{{2^2}}} + \frac{7}{{{2^3}}} \ldots .\infty$ is equal to

The sum of the series $2,5,8,11, \ldots$ is 60100 , then $n$ is equal to

The sum to $n$ terms of the series $\left( {{n^2} - {1^2}} \right) + 2\left( {{n^2} - {2^2}} \right) + 3\left( {{n^2} - {3^2}} \right) + \ldots ,$ is

A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments which are in AP. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. The value of 8 th instalment is