Let $a_1, a_2, a_3, \ldots$ be in an $A.P.$ such that $\sum_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1$,where $a_1 \neq 0$. If $\sum_{k=1}^{n} a_k = 0$,then $n$ is:

If the sum of $n$ terms of an arithmetic progression is $3n^2 + 5n$ and $T_m = 164$,then $m = \dots$

Let the positive integers be written in the form:
$1$
$2$ $3$
$4$ $5$ $6$
$7$ $8$ $9$ $10$
If the $k^{\text{th}}$ row contains exactly $k$ numbers for every natural number $k$,then the row in which the number $5310$ will be,is:

For any three positive real numbers $a, b, c$,if $9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$,then:

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:

Let the sequence $a_{n}$ be defined as follows:
$a_{1} = 1, a_{n} = a_{n-1} + 2$ for $n \ge 2$
Find the first five terms and write the corresponding series.