If ^nC_4, ^nC_5 and ^nC_6 are in arithmetic p | vedclass

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(A) Given that $^nC_4$, $^nC_5$, and $^nC_6$ are in $A$.$P$., we have $2(^nC_5) = ^nC_4 + ^nC_6$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$, we g

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$7$ or $14$

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(A) Given that $^nC_4$, $^nC_5$, and $^nC_6$ are in $A$.$P$., we have $2(^nC_5) = ^nC_4 + ^nC_6$.Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$, we get:$2 \cdot \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$.Dividing by $n!$ and multiplying by $6!(n-4)!$:$2 \cdot \frac{6!(n-4)!}{5!(n-5)!} = \frac{6!(n-4)!}{4!(n-4)!} + \frac{6!(n-4)!}{6!(n-6)!}$.$2 \cdot 6(n-4) = 6 \cdot 5 + (n-4)(n-5)$.$12n - 48 = 30 + n^2 - 9n + 20$.$n^2 - 21n + 98 = 0$.$(n-7)(n-14) = 0$.Thus, $n = 7$ or $n = 14$.

If $^nC_4$, $^nC_5$ and $^nC_6$ are in arithmetic progression ($A$.$P$.), then the value of $n$ is...

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