Question

Let $S=\{(a,b,c)\in N\times N\times N:a+b+c=21$, $a\le b\le c\}$ 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 $a=b=c=7[\because a\le b\le c]$
So, there are 7 such triplets.