The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + ........$ to $9$ terms is

Let $S_{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let, $S_{2}$ be the sum of first $4n$ terms of the same arithmetic progression. If $\left( S _{2}- S _{1}\right)$ is $1000,$ then the sum of the first $6 n$ terms of the arithmetic progression is equal to:

If the sum of the first $n$ terms of a series be $5{n^2} + 2n$, then its second term is

If $\left\{a_{i}\right\}_{i=1}^{n}$ where $n$ is an even integer, is an arithmetic progression with common difference $1$ , and $\sum \limits_{ i =1}^{ n } a _{ i }=192, \sum \limits_{ i =1}^{ n / 2} a _{2 i }=120$, then $n$ is equal to

If $\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c}$ are in an $A.P.$ and $\log _e \mathrm{a}-$ $\log _e 2 b, \log _e 2 b-\log _e 3 c, \log _e 3 c-\log _e a$ are also in an $A.P,$ then $a: b: c$ is equal to

The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is