The number of terms common to the two $A$.$P$.'s $3, 7, 11, \ldots, 407$ and $2, 9, 16, \ldots, 709$ 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 $(q-r)a + (r-p)b + (p-q)c = 0$.

The sum of the first $p, q,$ and $r$ terms of an $A.P.$ are $a, b,$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.

Let $a_{1}, a_{2}, a_{3}, \ldots$ be an $A.P.$ If $\frac{a_{1}+a_{2}+\ldots+a_{10}}{a_{1}+a_{2}+\ldots+a_{p}}=\frac{100}{p^{2}}, p \neq 10$,then $\frac{a_{11}}{a_{10}}$ is equal to :

Six numbers are in an $AP$ such that their sum is $3$. The first term is $4$ times the third term. Then,the fifth term is:

Let $9 < x_1 < x_2 < \ldots < x_7$ be in an $A.P.$ with common difference $d$. If the standard deviation of $x_1, x_2, \ldots, x_7$ is $4$ and the mean is $\overline{x}$,then $\overline{x} + x_6$ is equal to: