Find the sum of odd integers from $1$ to $2001 .$
The odd integers from $1$ to $2001$ are $1,3,5 \ldots \ldots .1999,2001$
This sequence forms an $A.P.$
Here, first term, $a=1$
Common difference, $d=2$
Here, $a+(n-1) d=2001$
$\Rightarrow 1+(n-1)(2)=2001$
$\Rightarrow 2 n-2=2000$
$\Rightarrow n=1001$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\therefore S_{n}=\frac{1001}{2}[2 \times 1+(1001-1) \times 2]$
$=\frac{1001}{2}[2+1000 \times 2]$
$=1001 \times 1001$
$=1002001$
Thus, the sum of odd numbers from $1$ to $2001$ is $1002001 .$
A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{{13}}{{12}}$, then the numbers are
Let $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},.....,$ $({x_i} \ne \,0\,for\,\,i\, = 1,2,....,n)$ be in $A.P.$ such that $x_1 = 4$ and $x_{21} = 20.$ If $n$ is the least positive integer for which $x_n > 50,$ then $\sum\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}}} \right)} $ is equal to.
Three numbers are in $A.P.$ whose sum is $33$ and product is $792$, then the smallest number from these numbers is
Write the first five terms of the sequences whose $n^{t h}$ term is $a_{n}=n \frac{n^{2}+5}{4}$