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If the co-efficient of $p^{th}, (p+1)^{th}$ and $(p+2)^{th}$ terms in the expansion of $(1+x)^n$ are in Arithmetic Progression (A.P.), then which one of the following is true?

  1. $n^2+4(4p+1)+4p^2-2=0$
  2. $n^2+4(4p+1)+4p^2+2=0$
  3. $(n-2p)^2=n+2$
  4. $(n+2p)^2=n+2$
in Numerical Ability by Boss (17.5k points)
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I think the options given are wrong because for $(1+x)^7$ the $1st (7),2nd(21)$ and $3rd(35)$ terms are in AP, but for p=1 and n=7 none of the options are getting satisfied.The correct answer should be $(n-2p)^2=5n-8p-2$
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Yes.
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How you solved?

1 Answer

+1 vote

We know that $(1+x)^n =\sum_{k\geq 0}^{n} \begin{pmatrix} n \\ k \end{pmatrix}x^n$ .

The expansion consists of $(n+1)$ terms, the coefficient of the $m^{th}$ term being $\begin{pmatrix} n \\ m-1 \end{pmatrix}$.

Since the coefficients of $p^{th}, (p+1)^{th}$ and $(p+2)^{th}$ terms are in A.P,

 

$\therefore$ $\begin{pmatrix} n\\ p-1 \end{pmatrix} + \begin{pmatrix} n \\ p+1 \end{pmatrix} = 2\begin{pmatrix} n \\ p \end{pmatrix}$

 

$\Rightarrow \frac{\not{n!}}{(n-p+1)! (p+1)!} + \frac{\not{n!}}{(n-p-1)!(p+1)!} = \frac{2\not{n!}}{(n-p)!p!}$

 

$\Rightarrow \frac{(p+1)p + (n-p+1)(n-p)}{(n-p+1)!(p+1)!} =\frac{2}{(n-p)!p!}$

 

$\Rightarrow \frac{(p+1)p + (n-p+1)(n-p)}{(n-p+1)(p+1)} = 2$

 

$\Rightarrow (p+1)p + (n-p+1)(n-p) = 2(n-p+1)(p+1)$

 

Solving this equation we get

 

$(n-2p)^2 =n+2$

 

Hence, the correct option is (C).

 

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