Sum of Terms of Arithmetic Progression

A person is to count $4500$ currency notes. Let $a_n$ denotes the number of notes he counts in the $n\mathrm{th}$ minute. If $a_1=a_2=\dots =a_{10}=150$ and $a_{10},a_{11},\dots$ are in $\mathrm{AP}$ with common difference $-2$, then the time taken by him to count all notes, is

A man saves $\mathrm{Rs}.200$ in each of the first three months of his service. In each of the subsequent months, his savings increases by $\mathrm{Rs}.40$ more than the savings of immediately previous month. His total saving from the start of service will be $\mathrm{Rs}.11040$ after

Let $a_1, a_2, a_3,\dots , a_{49}$ be in $\mathrm{A}.\mathrm{P}.$ such that$\sum _{k=0}^{12}{a}_{4k+1}=416$ and $a_9+a_{43}=66$. If $a_1^2+a_2^2+\dots +a_{17}^2=140m,$ then $m$ is equal to

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

For a positive integer $n,$ if the quadratic equation, $x(x+1)+(x+1)(x+2)+\dots +(x+\overline{n-1})(x+n)=10n$ has two consecutive integral solutions, then $n$ is equal to

Suppose that all the terms of an arithmetic progression ($\mathrm{A}.\mathrm{P}.$) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6:11$ and the seventh term lies in between $130$ and $140$, then the common difference of this $\mathrm{A}.\mathrm{P}.$ is

A pack contains $n$ cards numbered from $1$to $n.$ Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $1224.$ If the smaller of the numbers on the removed cards is $k,$ then $k-20 =\underline{ }$.

Let $a, b, c\in R.$ If$f\left(x\right)=a{x}^2+bx+c$ is such that $a+b+c=3$ and $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+xy,\forall x,y\in R,$ then $\sum _{n=1}^{10}f\left(n\right)$ is equal to

If the sum of $n$ terms of an A.P., is given by $S_n=a+bn+c{n}^2,$ where $a,b,c$ are independent of $n$, then

Statement 1: The sum of the series $1+\left(1+2+4\right)+\left(4+6+9\right)+\left(9+12+16\right)+...+\left(361+380+400\right)$is $8000.$

Statement 2: $\sum _{k=1}^n\left(k^3-{\left(k-1\right)}^3\right)=n^3,$ for any natural number $n.$

If $a_1,a_2,...,a_n$ are in A.P. with common difference $d\ne 0,$ then the sum of the series $\sin d \left[sec a_1 sec a_2+sec a_2 sec a_3+...+ sec a_{n-1} sec a_n\right]$ is

If the sum of $n$ terms of an A.P. is $cn\left(n-1\right)$, where $c\ne 0$, then the sum of the squares of these terms is

If $a_1,a_2,a_3,...,a_{2n+1}$ are in A.P., then $\frac{a_{2n+1}-a_1}{a_{2n+1}+a_1}+\frac{a_{2n}-a_2}{a_{2n}+a_2}+...+\frac{a_{n+2}-a_n}{a_{n+2}+a_n}$ is equal to

Concentric circles of radii $1,2,3,...,100 \mathrm{cm}$ are drawn. The interior of the smallest circle is coloured red and the angular regions are coloured alternately green and red, so that no two adjacent regions are of the same colour. Then, the total area to the green regions in sq. cm is equal

Let $a_1,a_2,a_3,....$ be terms of an A.P. If $\frac{a_1+a_2+...+a_p}{a_1+a_2+...+a_q}=\frac{p^2}{q^2},p\ne q$, then $\frac{a_6}{a_{21}}$equals

The first term of an arithmetic progression is $1$ and the sum of the first nine terms is $369.$ The first and the ninth term of a geometric progression coincide with the first and the ninth terms of the arithmetic progression. The value of the seventh term of the geometric progression is

The number of terms of an A.P., is even; the sum of the odd terms is $24$, and of the even terms is $30$, and the last term exceeds the first by $\frac{21}{2}$, then the number of terms in the series is

If $S_n$ denotes the sum of first $n$ terms of an A.P. and $\frac{S_{3n}-S_{n-1}}{S_{2n}-S_{2n-1}}=31$ ,then the value of $n$ is

In an A.P. of which $a$ is the first term, if the sum of the first $p$ terms is zero, then the sum of the next $q$ terms is

If the sum of $m$ terms of an A.P., is the same as the sum of its $n$ terms, then the sum of its $\left(m+n\right)$ terms is