If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

Let $x_1, x_2, x_3, x_4$ be in a geometric progression. If $2,7,9,5$ are subtracted respectively from $x_1, x_2, x_3, x_4$, then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24}\left(x_1 x_2 x_3 x_4\right)$ is:

If the sum of the first 20 terms of the series $\frac{4 \cdot 1}{4+3 \cdot 1^2+1^4}+\frac{4 \cdot 2}{4+3 \cdot 2^2+2^4}+\frac{4 \cdot 3}{4+3 \cdot 3^2+3^4}+\frac{4 \cdot 4}{4+3 \cdot 4^2+4^4}+\ldots \cdot$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where m and n are coprime, then $\mathrm{m}+\mathrm{n}$ is equal to :

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and $q$ respectively. Let d and D be the common differences of $\mathrm{AP}^{\prime} \mathrm{s}$ in $A$ and $B$ respectively such that $D=d+3, d>0$. If $\frac{p+q}{p-q}=\frac{19}{5}$, then $\mathrm{p}-\mathrm{q}$ is equal to