The sum of an infinite $G.P.$ with common ratio $r$ can be found if:

If the $12^{th}$ term of an $A.P.$ is $-13$ and the sum of the first four terms is $24,$ then what is the sum of the first $10$ terms?

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$, then the greatest number of the series will be:

The sum of the series $\frac{1}{{1 + {1^2} + {1^4}}} + \frac{2}{{1 + {2^2} + {2^4}}} + \frac{3}{{1 + {3^2} + {3^4}}} + \dots$ to $n$ 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

If $x, y, z$ are three positive numbers in $G.P.$,then $\frac{1 + \ln x}{2}, \frac{1 + \ln y}{4}, \frac{1 + \ln z}{8}$ are in