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Let $S=\{(a, b, c) \in N \times N \times N: a+b+c=21$, $a \leq b \leq c\}$ and $T=\{(a, b, c) \in N \times N \times N: a, b, c$ are in $A P \}$ where $N$ is the set of all natural numbers. Then, the number of elements in the set $\mathrm{S} \cap T$
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The correct answer is:
7
We have, $a+b+c=21$ and $\frac{a+c}{2}=b$ $\Rightarrow \quad a+c+\frac{a+c}{2}=21$
$\begin{array}{ll}\Rightarrow & a+c=14 \Rightarrow \frac{a+c}{2}=7 \\ \Rightarrow & b=7\end{array}$
So, a can take values from 1 to 6 . C can have values from 8 to 13
or a=b=c=7
So, there are 7 such triplets.
$\begin{array}{ll}\Rightarrow & a+c=14 \Rightarrow \frac{a+c}{2}=7 \\ \Rightarrow & b=7\end{array}$
So, a can take values from 1 to 6 . C can have values from 8 to 13
or a=b=c=7
So, there are 7 such triplets.
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