The sum of $n$ terms of two arithmetic progressions are in the ratio $(3n + 8) : (7n + 15)$. Find the ratio of their $12^{\text{th}}$ terms.

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 between $130$ and $140$,then the common difference of this $A.P.$ is

If $a, b, c$ are in Arithmetic Progression,then $(a - c)^2 = \dots$

Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}$,where $a > 2$. If $\alpha$ is such that $a, 4, \alpha, b$ are in $A$.$P$.,then the equation $\alpha x^2 - ax + 2(\alpha - 2b) = 0$ has :

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an arithmetic sequence are $a$,$b$,and $c$ respectively,then the value of $[a(q - r) + b(r - p) + c(p - q)]$ is:

If the sum and product of the first three terms in an $A.P.$ are $33$ and $1155$,respectively,then a value of its $11^{th}$ term is