The minimum value of ${\left( {\frac{3}{a} - 1} \right)^2} + {\left( {\frac{a}{b} - 1} \right)^2} + {\left( {\frac{b}{c} - 1} \right)^2} + {\left( {3c - 1} \right)^2}$ where $0 < a, b, c \leqslant 9$,is $p - q\sqrt{r}$; $p, q, r \in I$ and $q, r$ are coprime,then $(p + q + r)$ is equal to

  • A
    $16$
  • B
    $24$
  • C
    $27$
  • D
    $30$

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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 the $A.P.s$ in $A$ and $B$ respectively such that $D = d + 3$ and $d > 0$. If $\frac{p + q}{p - q} = \frac{19}{5}$,then $p - q$ is equal to:

If the sum of $n$ terms of an $A.P.$ is $3n^2 + 5n$ and $T_m = 164$,then $m = $

Let $S = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a+b+c=21, a \leq b \leq c\}$ and $T = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a, b, c \text{ are in } AP\}$, where $\mathbb{N}$ is the set of all natural numbers. Then, the number of elements in the set $S \cap T$ is:

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