Question

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 $AP \}$ where $N$ is the set of all natural numbers. Then, the number of elements in the set $S \cap T$ is

• A. 6
• B. 7
• C. 13
• D. 14

Answer 1

Answer: (B)

We have, $a+b+c=21$ and $\frac{a+c}{2}=b$
$\Rightarrow a+ c+\frac{a +c}{2}=21$
$\Rightarrow a +c=14$
$\Rightarrow \frac{a +c}{2}=7$
$\Rightarrow b=7$
So, a can take values from 1 to 6 ,
$c$ can have values from 8 to 13
or $a=b=c=7 [\because a \leq b \leq c]$
So, there are 7 such triplets.