Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$  If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals

Maximum value of sum of arithmetic progression $50, 48, 46, 44 ........$ is :-

If $1, \log _{10}\left(4^{x}-2\right)$ and $\log _{10}\left(4^{x}+\frac{18}{5}\right)$ are in
arithmetic progression for a real number $x$ then the value of the determinant $\left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{array}\right|$ is equal to ...... .

For $\mathrm{x} \geq 0$, the least value of $\mathrm{K}$, for which $4^{1+\mathrm{x}}+4^{1-\mathrm{x}}$, $\frac{\mathrm{K}}{2}, 16^{\mathrm{x}}+16^{-\mathrm{x}}$ are three consecutive terms of an $A.P.$ is equal to :

A number is the reciprocal of the other. If the arithmetic mean of the two numbers be $\frac{{13}}{{12}}$, then the numbers are

If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c, d$ are roots of $x^{2}-12 x+q=0$ where $a, b, c, d$ form a $G.P.$ Prove that $(q+p):(q-p)=17: 15$