Question

Let $S=\left\{\right(a,b,c)\in N\times N\times N:a+b+c=21,a\le b\le c\}$ and $T=\left\{\right(a,b,c)\in N\times N\times N:a,b,careinAP\}$, 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)

$7$

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 $6$ and $c$ can have values from $8$ to $13$
$a=b=c=7$
$[\because a\le b\le c]$
So, there are $7$ such triplets.