Write the first five terms of the following sequence and obtain the corresponding series :

$a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n\,>\,2$

The solution of the equation $(x + 1) + (x + 4) + (x + 7) + ......... + (x + 28) = 155$ is

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

If the sum of three numbers of a arithmetic sequence is $15$ and the sum of their squares is $83$, then the numbers are

Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$  If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals

Let $a_n, n \geq 1$, be an arithmetic progression with first term $2$ and common difference $4$ . Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum \limits_{n=1}^{10} M_n$ is