The first term of the $11^{th}$ group in the following sequence of groups $(1), (2, 3, 4), (5, 6, 7, 8, 9), \dots$ is:

The value of $\frac{1}{(1 + a)(2 + a)} + \frac{1}{(2 + a)(3 + a)} + \frac{1}{(3 + a)(4 + a)} + \dots + \infty$ is,(where $a$ is a constant).

There are $15$ terms in an arithmetic progression. Its first term is $5$ and their sum is $390$. The middle term is

If $a_1, a_2, a_3, ..., a_n$ are in $A.P.$,where $a_i > 0$ for all $i$,then the value of $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ... + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = $

If the harmonic mean between $a$ and $b$ is $H$,then the value of $\frac{1}{H - a} + \frac{1}{H - b}$ is

Let $b_1, b_2, \dots, b_n$ be a geometric sequence such that $b_1 + b_2 = 1$ and $\sum_{k=1}^{\infty} b_k = 2$. Given that $b_2 < 0$,then the value of $b_1$ is: