The first term of an $A.P.$ of consecutive integers is ${p^2} + 1$. The sum of $(2p + 1)$ terms of this series can be expressed as:

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$,then the greatest number of the series will be:

Consider the real-valued function $h: \{0, 1, 2, \ldots, 100\} \rightarrow \mathbb{R}$ such that $h(0) = 5$,$h(100) = 20$,and satisfying $h(p) = \frac{1}{2}\{h(p+1) + h(p-1)\}$ for every $p = 1, 2, \ldots, 99$. Then the value of $h(1)$ is:

Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1=15$,$27-2a_2 > 0$,and $a_k = 2a_{k-1} - a_{k-2}$ for $k = 3, 4, \ldots, 11$. If $\frac{a_1^2 + a_2^2 + \ldots + a_{11}^2}{11} = 90$,then the value of $\frac{a_1 + a_2 + \ldots + a_{11}}{11}$ is equal to

The $r$-th term of an arithmetic progression is $T_r$. Its first term is $a$ and the common difference is $d$. If for some positive integers $m, n, m \neq n,$ we have $T_m = 1/n$ and $T_n = 1/m,$ then $a - d = \dots\dots.$

If $a + 2b + 3c = 6$,then the greatest value of $abc^2$ is (where $a, b, c$ are positive real numbers).