Let $3, a, b, c$ be in $A.P.$ and $3, a-1, b+1, c+9$ be in $G.P.$ Then, the arithmetic mean of $a, b$ and $c$ is :

If three unequal numbers $p,\;q,\;r$ are in $H.P.$ and their squares are in $A.P.$, then the ratio $p:q:r$ is

The sum of three numbers in $G.P.$ is $56.$ If we subtract $1,7,21$ from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Three numbers form a $G.P.$ If the ${3^{rd}}$ term is decreased by $64$, then the three numbers thus obtained will constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$, a $G.P.$ will be formed again, then the numbers will be

If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ such that $ a <  b$ $ < c$ and $a+b+c\,= \frac{3}{4}$ , then the value of $a$ is

If $a,\;b,\;c$ are in $A.P.$ and $a,\;c - b,\;b - a$ are in $G.P. $ $(a \ne b \ne c)$, then $a:b:c$ is