Let $S_1, S_2, \dots, S_{101}$ be the consecutive terms of an $A.P$. If $\frac{1}{S_1 S_2} + \frac{1}{S_2 S_3} + \dots + \frac{1}{S_{100} S_{101}} = \frac{1}{6}$ and $S_1 + S_{101} = 50$,then $|S_1 - S_{101}|$ is equal to

If $a, b, c$ are in Arithmetic Progression,then $\frac{1}{\sqrt{b} + \sqrt{c}}, \frac{1}{\sqrt{c} + \sqrt{a}}, \frac{1}{\sqrt{a} + \sqrt{b}}$ are in:

The sum of the first $n$ natural numbers is

Jairam purchased a house for Rs. $15000$ and paid Rs. $5000$ at once. He promised to pay the remaining amount in annual installments of Rs. $1000$ with $10\%$ per annum interest on the outstanding balance. How much total money will be paid by Jairam in Rs.?

If the $7^{th}$ term of an Arithmetic Progression $(AP)$ is $40$,then the sum of its first $13$ terms is........

Let the sum of the first $n$ terms of a non-constant $A.P.$,$a_1, a_2, a_3, \dots$ be $S_n = 50n + \frac{n(n - 7)}{2}A$,where $A$ is a constant. If $d$ is the common difference of this $A.P.$,then the ordered pair $(d, a_{50})$ is equal to