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 :

Find the $17^{\text {th }}$ and $24^{\text {th }}$ term in the following sequence whose $n^{\text {th }}$ term is $a_{n}=4 n-3$

If the ${p^{th}}$ term of an $A.P.$ be $\frac{1}{q}$ and ${q^{th}}$ term be $\frac{1}{p}$, then the sum of its $p{q^{th}}$ terms will be

For three positive integers $p , q , r , x ^{ pq p ^2}= y ^{ qr }= z ^{ p ^2 r }$ and $r=p q+1$ such that $3,3 \log _y x, 3 \log _z y, 7 \log _x z$ are in A.P. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to

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 $(q+p):(q-p)=17: 15$

If $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$ be the $A.M.$ of $a$ and $b$, then $n=$