If the coefficients of p th - p -1th and p -2th terms in the expansion of 1- x n are in A-P- thenA- n2-n4 p-1-4 p2-2-0B- n2-2 n p-4 p2-0C- n2-n4 p-1-4 p2-0D- None of these

The correct option is A n^2 n(4p + 1) + 4p^2 2 = 0 2^nc_p=n_c_p 1+p_c_p+1 ⇒2/(n p)!P!=1/(n p+1)!(p 1)!+1/(n p 1)!(p+1)! ⇒2/(n p)p=1/(n p+1)(n p)+1/(p+1)p ...

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Byju's Answer

Standard XI

Mathematics

Binomial Theorem

If the coeffi...

Question

If the coefficients of $p^{t h}$ , $(p + 1)^{t h}$ and $(p + 2)^{t h}$ terms in the expansion of $(1 + x)^{n}$ are in A.P., then

$n^{2} - 2 n p + 4 p^{2}$ = 0

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$n^{2} - n (4 p + 1) + 4 p^{2} - 2$ = 0

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$n^{2} - n (4 p + 1) + 4 p^{2}$ = 0

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None of these

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Solution

The correct option is B
$n^{2} - n (4 p + 1) + 4 p^{2} - 2$ = 0

2 $^{n} c_{p} = n_{c_{p - 1}} + p_{c_{p + 1}} \Rightarrow \frac{2}{(n - p)! P!} = \frac{1}{(n - p + 1)! (p - 1)!} + \frac{1}{(n - p - 1)! (p + 1)!} \Rightarrow \frac{2}{(n - p) p} = \frac{1}{(n - p + 1) (n - p)} + \frac{1}{(p + 1) p} o n s i m p l i f i n g w e g e t n^{2} - n (4 p + 1) + 4 p^{2} - 2 = 0$

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Similar questions

If the coefficient of $m^{t h}$ , $(m + 1)^{t h}$ and $(m + 2)^{t h}$ terms in the expansion of $(1 + x)^{n}$ are in A.P., then:

Q. If the coefficient of

r^{t h}, (r + 1)^{t h}

and

(r + 2)^{t h}

terms in the expansion of

(1 + x)^{n}

are in A.P., then show that

n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0

Q. If the coefficients of

r^{t h}, {(r + 1)}^{t h}, {(r + 2)}^{t h}

terms in

{(1 + x)}^{n}

are in A.P. then show that

n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0

Q. If the coefficient of

r^{t h}, (r + 1)^{t h}

and

(r + 2)^{t h}

terms in the binomial expansion of

{(1 + y)}^{n}

are in A.P, then

n^{2} - n (4 r + 1) + {4 r}^{2} - 2

is equal to

Q. In the expansion of

{(1 + x)}^{n}

, the coefficients of

p^{t h}

and

(p + 1)^{t h}

terms are respectively

p

and

q

, then

p + q =

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If the coefficients of pth, (p+1)th and (p+2)th terms in the expansion of (1+x)n are in A.P., then

The correct option is B n2−n(4p+1)+4p2−2 = 0 2ncp=ncp−1+pcp+1⇒2(n−p)!P!=1(n−p+1)!(p−1)!+1(n−p−1)!(p+1)!⇒2(n−p)p=1(n−p+1)(n−p)+1(p+1)pon simplifing we getn2−n(4p+1)+4p2−2=0

If the coefficients of $p^{t h}$ , $(p + 1)^{t h}$ and $(p + 2)^{t h}$ terms in the expansion of $(1 + x)^{n}$ are in A.P., then