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

Let ${a_1}, {a_2}, \dots, {a_{49}}$ be in $A.P.$ such that $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $\sum_{r = 1}^{17} a_r^2 = 140m$,then $m = \dots$

If the sum of the first $21$ terms of the series $\log _{9^{1 / 2}} x + \log _{9^{1 / 3}} x + \log _{9^{1 / 4}} x + \dots$ where $x > 0$ is $504$,then $x$ is equal to:

$A$ man deposited $Rs. 10000$ in a bank at the rate of $5\%$ simple interest annually. Find the amount in the $15^{\text{th}}$ year since he deposited the amount and also calculate the total amount after $20$ years.

The number of terms in the series $101 + 99 + 97 + \dots + 47$ is

If the sum of the first $10$ terms of an arithmetic progression is $4$ times the sum of its first $5$ terms,then the ratio of its first term to the common difference is......