Maximum value of the sum of the arithmetic progression $50, 48, 46, 44, \dots$ is:

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$,then $8 \sigma^2$ is equal to

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

Let $a_{1}, a_{2}, \ldots, a_{21}$ 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 $A.P.$ is $189$,then $a_{6} a_{16}$ is equal to:

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 $\sum_{n=1}^{10} M_n$ is

The $p^{\text{th}}$,$q^{\text{th}}$,and $r^{\text{th}}$ terms of an $A.P.$ are $a$,$b$,and $c$ respectively. Show that $(q-r)a + (r-p)b + (p-q)c = 0$.