Question

Let $S = \{ (a,b,c) \in N \times N \times N{\text{ }}:{\text{ }}a + b + c = 21,{\text{ }}a \leqslant b \leqslant c\} $ and $T = \{ (a,b,c) \in $
$N \times N \times N:{\text{ }}a,b,c{\text{ are in }}AP{\text{\} , }}$where $N$is the set of all natural numbers.
Then, the number of elements in the set $S \cap T$ is:
$(A){\text{ }}$6
$(B){\text{ }}$7
$(C){\text{ }}$13
$(D){\text{ }}$14

Answer 1

Hint: Find the middle term(b) of A.P. So, that number of variables reduces to two. And we can form a triplet of 3 numbers in A.P by changing first and last numbers.

Complete step-by-step answer:

As we are given that,
$a,b,c$ are in A.P and
$ \Rightarrow a + b + c = 21{\text{ }}$ (1)

So, from the condition of A.P that if three numbers are in A.P then,
Middle number will be equal to the half of sum of first and last numbers. So,
$ \Rightarrow \dfrac{{a + c}}{2} = b$

Now, putting the value of b in equation 1. We get,
$ \Rightarrow a + \dfrac{{a + c}}{2} + c = 21$

On solving the above equation. We get,
$
   \Rightarrow a + c = 14 \\
   \Rightarrow {\text{So, }}b = \dfrac{{a + c}}{2} = 7 \\
 $

Now, as we know that it is given that,
$ \Rightarrow (a,b,c) \in N,{\text{ }}a \leqslant b \leqslant c,{\text{ }}a + b + c = 21$

So, a can take values from 1 to 7.
And. c can take values from 7 to 13.

So, there will be 7 triplets possible. And they will be,
$ \Rightarrow {\text{\{ 1,7,13\} , \{ 2,7,12\} , \{ 3,7,11\} , \{ 4,7,10\} , \{ 5,7,9\} , \{ 6,7,8\} and \{ 7,7,7\} }}$
$ \Rightarrow $ Hence, the correct answer will be B.

Note: In such types of problems first, we should find the second term of the A.P using condition${\text{ }}b = \dfrac{{a + c}}{2}$. And then we can change the value of the first and third term of that A.P. according to the given conditions. As this will be the easiest and efficient way to reach the required answer.