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

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

If the $9^{th}$ term of an $A.P.$ be zero, then the ratio of its $29^{th}$ and $19^{th}$ term is

If $a_1, a_2, a_3 …………$ an are in $A.P$ and $a_1 + a_4 + a_7 + …………… + a_{16} = 114$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

Let $A =\left\{1, a _{1}, a _{2} \ldots \ldots a _{18}, 77\right\}$ be a set of integers with $1< a _{1}< a _{2}<\ldots \ldots< a _{18}<77$. Let the set $A + A =\{ x + y : x , y \in A \} \quad$ contain exactly $39$ elements. Then, the value of $a_{1}+a_{2}+\ldots \ldots+a_{18}$ is equal to...........

Suppose that all the terms of an arithmetic progression ($A.P.$) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6: 11$ and the seventh term lies in between $130$ and $140$ , then the common difference of this $A.P.$ is

If $\left\{a_{i}\right\}_{i=1}^{n}$ where $n$ is an even integer, is an arithmetic progression with common difference $1$ , and $\sum \limits_{ i =1}^{ n } a _{ i }=192, \sum \limits_{ i =1}^{ n / 2} a _{2 i }=120$, then $n$ is equal to