If $S_1, S_2$ and $S_3$ are the sums of the first $n_1, n_2$ and $n_3$ terms of an arithmetic progression respectively,then $\frac{S_1}{n_1}(n_2 - n_3) + \frac{S_2}{n_2}(n_3 - n_1) + \frac{S_3}{n_3}(n_1 - n_2) = ....$

Between $1$ and $31$,$m$ numbers have been inserted in such a way that the resulting sequence is an $A.P.$ and the ratio of the $7^{\text{th}}$ and $(m-1)^{\text{th}}$ inserted numbers is $5:9$. Find the value of $m$.

The first term of an $A$.$P$. of $30$ non-negative terms is $\frac{10}{3}$. If the sum of this $A$.$P$. is the cube of its last term,then its common difference is:

$A$ man saves $200$ rupees in each of the first three months of his job. In the subsequent months,his savings increase by $40$ rupees compared to the previous month. After how many months from the start of the job will his total savings be $11040$ rupees?

Let $S = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a+b+c=21, a \leq b \leq c\}$ and $T = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a, b, c \text{ are in } AP\}$,where $\mathbb{N}$ is the set of all natural numbers. Then,the number of elements in the set $S \cap T$ is:

If the $n^{th}$ terms of two $A.P.$'s are $3n + 8$ and $7n + 15$,then the ratio of their $12^{th}$ terms will be: