If the sum of three consecutive terms of an $A.P.$ is $51$ and the product of last and first term is $273$, then the numbers are

Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. If the sum of this AP is $189,$ then  $a_{6} \mathrm{a}_{16}$ is equal to :

If the ${p^{th}},\;{q^{th}}$ and ${r^{th}}$ term of an arithmetic sequence are $a , b$ and $c$ respectively, then the value of $[a(q – r)$ + $b(r – p)$ $ + c(p – q)] = $

If $a_1, a_2, a_3, …….$ are in $A.P.$ such that $a_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P.$ is

If $x=\sum \limits_{n=0}^{\infty} a^{n}, y=\sum\limits_{n=0}^{\infty} b^{n}, z=\sum\limits_{n=0}^{\infty} c^{n}$, where $a , b , c$ are in $A.P.$ and $|a| < 1,|b| < 1,|c| < 1$, $abc \neq 0$, then