For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then
$a,b,c$ are in $G.P.$
$b,c,a$ are in $G.P.$
$b,c,a$ are in $A.P.$
$a,b,c$ are in $A.P.$
The number of terms in the series $101 + 99 + 97 + ..... + 47$ is
If the sum of first $11$ terms of an $A.P.$, $a_{1} a_{2}, a_{3}, \ldots$is $0\left(\mathrm{a}_{1} \neq 0\right),$ then the sum of the $A.P.$, $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1},$ where $k$ is equal to
$150$ workers were engaged to finish a piece of work in a certain number of days. $4$ workers dropped the second day, $4$ more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is
If $\frac{1}{{b - c}},\;\frac{1}{{c - a}},\;\frac{1}{{a - b}}$ be consecutive terms of an $A.P.$, then ${(b - c)^2},\;{(c - a)^2},\;{(a - b)^2}$ will be in
Let the digits $a, b, c$ be in $A.P.$ Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in $A.P.$ at least once. How many such numbers can be formed?