The sum $1+\frac{1+3}{2!}+\frac{1+3+5}{3!}+\frac{1+3+5+7}{4!}+\ldots$ upto $\infty$ terms, is equal to

Let $a_1, a_2, a_3, \ldots$. be a G.P. of increasing positive numbers. If $a_3 a_5=729$ and $a_2+a_4=\frac{111}{4}$, then $24\left(a_1+a_2+a_3\right)$ is equal to

The sum $1+3+11+25+45+71+\ldots$ upto 20 terms, is equal to

The number of terms of an A.P. is even; the sum of all the odd terms is 24 , the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the A.P. is :