Jitender Gupta and Divya Malik Solutions for Exercise 6: CHALLENGERS

Five distinct positive integers are in arithmetic progression with a positive common difference. If their sum is $10020$, then the smallest possible value of the last term is 

The sum of the series $1+\left(1+2\right)+\left(1+2+3\right)+\left(1+2+3+4\right)+....+\left(1+2+3+...+20\right)$ is

An $AP$ starts with a positive fraction and every alternate term is an integer. If the sum of the first $11$ terms is $33$, then the fourth term is:

The sum of the series

${45}^2-{43}^2+{44}^2-{42}^2+{43}^2-{41}^2+{42}^2-{40}^2+...$ up to $30$ terms is

Take a point $A\left(3,4\right)$ on the graph and draw two lines from it, one is parallel to $x$- axis and another parallel to $y$-axis. Again, take four points on both lines on both side of $A$, such that their $x$-coordinates and $y$-coordinates form an AP with common difference $2$. Then, the area of the circle, passing  through these four points, is

If $a,b,c$ and $d$ are in GP, then $\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)$ is equal to-

If $a,b$ and $c$ respectively the $p$th, $q$ th and $r$ th terms of a GP, then the value of 
$\left(q-r\right)\log a+\left(r-p\right)\log b+\left(p-q\right)\log c$is