If n C 4- n C 5- n C 6 are in A-P- then the values of n is-areA- 7B- 14C- 8D- 16

The correct options are A 7 B 14As nbsp;^nC_4, ^nC_5, ^nC_6 are in A.P., so ^nC_4+ ^nC_6 = 2 ^nC_5 ⇒ 1+ ^nC_6 ^nC_4 = 2× ^nC_5 ^nC_4 ⇒ 1+n!4!(n 4)!n!6!(n ...

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Byju's Answer

Standard XII

Mathematics

Pascal Triangle

If n C 4, n C...

Question

If $^{n} C_{4},^{n} C_{5},^{n} C_{6}$ are in A.P., then the value(s) of $n$ is/are

7

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14

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8

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Solution

The correct options are
A $7$
B $14$
As $^{n} C_{4},^{n} C_{5},^{n} C_{6}$ are in A.P., so
$^{n} C_{4} +^{n} C_{6} = 2^{n} C_{5} \Rightarrow 1 + \frac{^{n} C_{6}}{^{n} C_{4}} = 2 \times \frac{^{n} C_{5}}{^{n} C_{4}} \Rightarrow 1 + \frac{n! 4! (n - 4)!}{n! 6! (n - 6)!} = 2 \times \frac{n! 4! (n - 4)!}{5! (n - 5)! n!} \Rightarrow 1 + \frac{(n - 4) (n - 5)}{6 \times 5} = 2 \times \frac{(n - 4)}{5} \Rightarrow 30 + n^{2} - 9 n + 20 = 12 n - 48 \Rightarrow n^{2} - 21 n + 98 = 0 \Rightarrow (n - 7) (n - 14) = 0 ∴ n = 7, 14$

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If nC4, nC5, nC6 are in A.P., then the value(s) of n is/are

The correct options are A 7 B 14As nC4, nC5, nC6 are in A.P., so nC4+ nC6=2 nC5⇒1+ nC6 nC4=2× nC5 nC4⇒1+n!4!(n−4)!n!6!(n−6)!=2×n!4!(n−4)!5!(n−5)!n!⇒1+(n−4)(n−5)6×5=2×(n−4)5⇒30+n2−9n+20=12n−48⇒n2−21n+98=0⇒(n−7)(n−14)=0∴n=7,14

If $^{n} C_{4},^{n} C_{5},^{n} C_{6}$ are in A.P., then the value(s) of $n$ is/are