Question

If a, b, c are three natural numbers in AP and a+b+c=21 then the possible number of values of the ordered triplet $(a,b,c)$ is

• A. 15
• B. 14
• C. 13
• D. none of these

Answer 1

The correct option is C 13

We know that
$b=\frac{a+c}{2}$ since they are in A.P
Hence $2b=a+c$
$2(a+b+c)=42$
$2a+2c+2b=42$
$2a+2c+a+c=42$
$3(a+c)=42$
$a+c=14$
Therefore $b=7$
Now $a,b,cϵN$, we get
$a=1,c=13$,
$a=2,c=12$
:
:
:
$a=13,c=1$
Hence we will have 13 ordered pairs of $(a,b,c)$ as $a=(1,2,...13)$
$c=(13,12,\mathrm{11...1})$ and $b=7$