If n C 4- n C 5 and n C 6 are in A-P- then n can be -A- 11B- 14C- 12D- 9

The correct option is B 14^nC_4, ^nC_5, ^nC_6 are in A.P. ∴ 2× ^nC_5= ^nC_4+ ^nC_6 ⇒ 2×n!5!× (n 5)!=n!4!× (n 4)!+n!6!× (n 6)! ⇒25(n 5)=1(n 4)(n 5)+130 From ...

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Byju's Answer

Standard XI

Mathematics

Definition of Functions

If n C 4, n C...

Question

If $^{n} C_{4},^{n} C_{5}$ and $^{n} C_{6}$ are in A.P., then $n$ can be :

9

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Solution

The correct option is D $14$
$^{n} C_{4},^{n} C_{5},^{n} C_{6}$ are in A.P.

$∴ 2 \times^{n} C_{5} =^{n} C_{4} +^{n} C_{6}$
$\Rightarrow 2 \times \frac{n!}{5! \times (n - 5)!} = \frac{n!}{4! \times (n - 4)!} + \frac{n!}{6! \times (n - 6)!}$

$\Rightarrow \frac{2}{5 (n - 5)} = \frac{1}{(n - 4) (n - 5)} + \frac{1}{30}$

From options, $n = 14$ satisfies the above equation.

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If nC4, nC5 and nC6 are in A.P., then n can be :

The correct option is D 14 nC4, nC5, nC6 are in A.P. ∴2× nC5= nC4+ nC6 ⇒2×n!5!×(n−5)!=n!4!×(n−4)!+n!6!×(n−6)! ⇒25(n−5)=1(n−4)(n−5)+130 From options, n=14 satisfies the above equation.

If $^{n} C_{4},^{n} C_{5}$ and $^{n} C_{6}$ are in A.P., then $n$ can be :