Find the $9^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=(-1)^{n-1} n^{3}$

In an $A.P.,$ the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{th}$ term is $-112$

The sum of first $n$ natural numbers is

Consider a sequence whose sum of first $n$ -terms is given by $S_n = 4n^2 + 6n, n \in N$, then $T_{15}$ of this sequence is -

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

Let $S_n$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$. . If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then $\mathrm{n}$ equals