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If the coefficients of $r^{th},\left ( r+1 \right )^{th}\: and \: \left ( r+2 \right )^{th}$

terms in the binomial expansion of $\left ( 1+y \right )^{m}$ are in A.P., then $m$ and $r$ satisfy the equation

Option 1)
$m^{2}-m\left ( 4r-1 \right )+4r^{2}+2= 0$

Option 2)
$m^{2}-m\left ( 4r+1 \right )+4r^{2}-2= 0$

Option 3)
$m^{2}-m\left ( 4r+1 \right )+4r^{2}+2= 0$

Option 4)
$m^{2}-m\left ( 4r-1 \right )+4r^{2}-2= 0$

Answers (1)

best_answer

As we learnt in

Arithmetic mean of two numbers (AM) -

$A=\frac{a+b}{2}$

- wherein

It is to be noted that the sequence a, A, b, is in AP where, a and b are the two numbers.

Given that:

T_r, T_r+1, T_r+2in A.P

$2\times T_{r+1}=T_{r}+T_{r+2}$

$2\times\ ^mC_{r}=\ ^mC_{r-1}+\ ^mC_{r+1}$

$\Rightarrow \frac{2\times m!}{r!(m-r)!}=\frac{m!}{(r-1)!(m-r+1)!}+\frac{m!}{(r+1)!(m-r-1)!}$

$\Rightarrow \frac{2}{r!(m-r)!}=\frac{1}{(r-1)!(m-r+1)!}+\frac{1}{(r+1)!(m-r-1)!}$

$\Rightarrow \frac{2}{r(r-1)!(m-r)(m-r-1)!}=\frac{1}{(r-1)!(m-r+1)(m-r)!}+\frac{1}{(r+1)r(r-1)!(m-r-1)!}$

$\Rightarrow \frac{2}{r(m-r)}=\frac{1}{(m-r+1)}=\frac{1}{r(r+1)}$

$\Rightarrow \frac{1}{m-r}[\frac{2}{r}-\frac{1}{(m-r+1)}]=\frac{1}{r(r+1)}$

$\Rightarrow \frac{1}{m-r}\left[\frac{2m-3r+2}{r(m-r+1)} \right ]=\frac{1}{r(r+1)}$

$\Rightarrow$ (2m-3r+2)(r+1)=(m-r)(m-r+1)

$\Rightarrow$ 2m(r+1)-3r(r+1)+2(r+1)=(m-r)²+(m-r)

$\Rightarrow$ 2mr+2m-3r²+2=m²-2mr+r²+m

$\Rightarrow$ m²-m(4r+1)+4r²-2=0

Option 1)

$m^{2}-m\left ( 4r-1 \right )+4r^{2}+2= 0$

This is incorrect

Option 2)

$m^{2}-m\left ( 4r+1 \right )+4r^{2}-2= 0$

This is correct

Option 3)

$m^{2}-m\left ( 4r+1 \right )+4r^{2}+2= 0$

This is incorrect

Option 4)

$m^{2}-m\left ( 4r-1 \right )+4r^{2}-2= 0$

This is incorrect

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