If $\frac{{3 + 5 + 7 + ..........{\rm{to}}\;n\;{\rm{terms}}}}{{5 + 8 + 11 + .........{\rm{to}}\;10\;{\rm{terms}}}} = 7$, then the value of $n$ is

The sum of the first four terms of an $A.P.$ is $56 .$ The sum of the last four terms is $112.$ If its first term is $11,$ then find the number of terms.

If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.,$ prove that $a, b, c$ are in $A.P.$

Let ${a_1},{a_2},.......,{a_{30}}$ be an $A.P.$, $S = \sum\limits_{i = 1}^{30} {{a_i}} $ and $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $.If ${a_5} = 27$ and $S - 2T = 75$ , then $a_{10}$ is equal to

The number of terms in the series $101 + 99 + 97 + ..... + 47$ is

If ${a_1},\,{a_2},....,{a_{n + 1}}$ are in $A.P.$, then $\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ..... + \frac{1}{{{a_n}{a_{n + 1}}}}$ is