Question

Let $S=\left\{\right(a,b,c)ϵN\times N\times N:a+b+c=21.a\le b\le c\}$ and $T=\{a,b,c)ϵN\times N\times N:a,b,c,areinA.P.\}$, 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

The correct option is B $7$

$a+b+c=21$ and $b=\frac{a+c}{2}$
$\Rightarrow a+c=14$ and $b=7$
So, a can take values from $1$ to $6$, when $c$ ranges from $13$ to $8$, or $a=b=c=7$
So, $7$ triplets