Question
If the coefficients of $${r^{th}},{\left( {r + 1} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the the binomial expansion of $${\left( {1 + y} \right)^{m}}$$ are in A.P., then $$m$$ and $$r$$ satisfy the equation
A.
$${m^2} - m\left( {4r - 1} \right) + 4{r^2} - 2 = 0$$
B.
$${m^2} - m\left( {4r + 1} \right) + 4{r^2} + 2 = 0$$
C.
$${m^2} - m\left( {4r + 1} \right) + 4{r^2} - 2 = 0$$
D.
$${m^2} - m\left( {4r - 1} \right) + 4{r^2} + 2 = 0$$
Answer :
$${m^2} - m\left( {4r + 1} \right) + 4{r^2} - 2 = 0$$
Solution :
$$\eqalign{
& {\text{Given,}}{{\text{ }}^m}{C_{r - 1}},{\,^m}{C_r},{\,^m}{C_{r + 1}}\,{\text{are in A}}{\text{.P}}{\text{.}} \cr
& {{\text{2}}^m}{C_r} = {\,^m}{C_{r - 1}} + {\,^m}{C_{r + 1}} \cr
& \Rightarrow 2 = \frac{{^m{C_{r - 1}}}}{{^m{C_r}}} + \frac{{^m{C_{r + 1}}}}{{^m{C_r}}} = \frac{r}{{m - r + 1}} + \frac{{m - r}}{{r + 1}} \cr
& \Rightarrow {m^2} - m\left( {4r + 1} \right) + 4{r^2} - 2 = 0. \cr} $$