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If a, b, c are three natural numbers in AP and a + $\mathrm{b}+\mathrm{c}=21$ then the possible number of values of the ordered triplet $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ is
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The correct answer is:
13
$\quad$ Let $a=b-d$ and $c=b+d$, then $a+b+c=21$ $\Rightarrow \mathrm{b}=7$
So, the equation is $\mathrm{a}+\mathrm{c}=14$
$\therefore$ No. of solution $=$ coeff. of $x^{14}$ in $\left(x+x^{2}+\ldots .\right)$ $={ }^{13} \mathrm{C}_{12}=13$
So, the equation is $\mathrm{a}+\mathrm{c}=14$
$\therefore$ No. of solution $=$ coeff. of $x^{14}$ in $\left(x+x^{2}+\ldots .\right)$ $={ }^{13} \mathrm{C}_{12}=13$
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