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

Answer: (C)

Let $a=b-d$ and $c=b+d$, then $a+b+c=21$
$\Rightarrow b=7$
So, the equation is $a+c=14$
$\therefore$ No. of solution $=$ coeff. of $x^{14}$ in $\left(x+x^2+\dots .\right)$
$={}^{13}{C}_{12}=13$