G Tewani Solutions for Chapter: Progression and Series, Exercise 2: Archives

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

If ${\left(10\right)}^9+2{\left(11\right)}^1{\left(10\right)}^8+3{\left(11\right)}^2{\left(10\right)}^7+\dots +10{\left(11\right)}^9=k{\left(10\right)}^9$, then $k$ is equal to

The sum of first $9$ terms of the series $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots$ is

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

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^2+2·2^2+3^2+2·4^2+5^2+2·6^2+\dots \dots$ If $B-2A=100\lambda ,$ then $\mathrm{\lambda }$ is equal to

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

Let $a_1, a_2, a_3,\dots .$ be in harmonic progression with $a_1=5$ and $a_{20}=25$. The least positive integer $n$ for which $a_n<0$ is