If the coefficient of 3rd- 4th and 5th terms in the expansion of 1-xn- n - are in A-P- then the number of possible values of n is

T_3=^nC_2· x^2 T_4=^nC_3· x^3 T_5=^nC_4· x^4 According to condition, 2×^nC_3=^nC_2+^nC_4 ⇒ 2 = ^nC_2 ^nC_3+ ^nC_4 ^nC_3 ⇒3n 2+n 34=2 ⇒ 12+n^2 5n+6 = 8n 16 ⇒ ...

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Standard XII

Mathematics

Binomial Theorem for Any Index

If the coeffi...

Question

If the coefficient of $3^{rd}, 4^{th}$ and $5^{th}$ terms in the expansion of $(1 + x)^{n}, n \in N$ are in A.P., then the number of possible value(s) of $n$ is

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Solution

The correct option is A $0$
$T_{3} =^{n} C_{2} \cdot x^{2}$
$T_{4} =^{n} C_{3} \cdot x^{3}$
$T_{5} =^{n} C_{4} \cdot x^{4}$
According to condition,
$2 \times^{n} C_{3} =^{n} C_{2} +^{n} C_{4} \Rightarrow 2 = \frac{^{n} C_{2}}{^{n} C_{3}} + \frac{^{n} C_{4}}{^{n} C_{3}} \Rightarrow \frac{3}{n - 2} + \frac{n - 3}{4} = 2 \Rightarrow 12 + n^{2} - 5 n + 6 = 8 n - 16 \Rightarrow n^{2} - 13 n + 34 = 0 \Rightarrow n = \frac{13 \pm \sqrt{169 - 136}}{2} \Rightarrow n = \frac{13 \pm \sqrt{33}}{2}$
As $n \in N$ , so no value of $n$ is possible.

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Binomial Theorem for Any Index

Standard XII Mathematics

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If the coefficient of 3rd,4th and 5th terms in the expansion of (1+x)n,n∈N are in A.P., then the number of possible value(s) of n is

The correct option is A 0T3=nC2⋅x2 T4=nC3⋅x3 T5=nC4⋅x4 According to condition, 2×nC3=nC2+nC4⇒2= nC2 nC3+ nC4 nC3⇒3n−2+n−34=2⇒12+n2−5n+6=8n−16⇒n2−13n+34=0⇒n=13±√169−1362⇒n=13±√332 As n∈N, so no value of n is possible.

If the coefficient of $3^{rd}, 4^{th}$ and $5^{th}$ terms in the expansion of $(1 + x)^{n}, n \in N$ are in A.P., then the number of possible value(s) of $n$ is