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Step-by-Step Solution
Step 1: Identify the General Term in the Binomial Expansion
The general term of the binomial expansion of $(1 + y)^m$ is given by
$T_{k+1} = \binom{m}{k} \, y^k,$
where $\binom{m}{k}$ is the binomial coefficient.
Step 2: Recognize the Coefficients in an Arithmetic Progression
The question states that the coefficients of the $r^\text{th}$, $(r+1)^\text{th}$, and $(r+2)^\text{th}$ terms are in Arithmetic Progression (A.P.).
Let those coefficients be $C_r = \binom{m}{r-1}$, $C_{r+1} = \binom{m}{r}$, and $C_{r+2} = \binom{m}{r+1}$ respectively.
For three numbers $a, b, c$ to be in A.P., we must have
$$2b = a + c.$$
Hence, the condition becomes
$$2 \binom{m}{r} = \binom{m}{r-1} + \binom{m}{r+1}.$$
Step 3: Use an Illustrative Example (Setting $r=2$)
As a simpler illustration (common approach in many problems), take $r = 2$:
The coefficients of the 2nd, 3rd, and 4th terms of $(1 + y)^m$ must be in A.P.
These coefficients are $\binom{m}{1}$, $\binom{m}{2}$, and $\binom{m}{3}$ respectively.
Hence, the A.P. condition for these coefficients gives:
$$2 \binom{m}{2} = \binom{m}{1} + \binom{m}{3}.$$
Step 4: Express Binomial Coefficients in Factorial Form
Recall that
$$\binom{m}{k} = \frac{m \cdot (m - 1) \cdot \ldots \cdot (m - k + 1)}{k!}.$$
Therefore:
$\binom{m}{1} = m,$
$\binom{m}{2} = \frac{m(m-1)}{2},$
$\binom{m}{3} = \frac{m(m-1)(m-2)}{6}.$
Step 5: Substitute and Simplify
Substitute these into the A.P. condition:
$$2 \left(\frac{m(m-1)}{2}\right) = m + \frac{m(m-1)(m-2)}{6}.$$
Simplify step-by-step:
Left side: $2 \cdot \frac{m(m-1)}{2} = m(m-1).$
Right side: $m + \frac{m(m-1)(m-2)}{6}.$
So the equation becomes
$$m(m-1) = m + \frac{m(m-1)(m-2)}{6}.$$
Multiply through by 6 to clear the fraction:
$$6m(m-1) = 6m + m(m-1)(m-2).$$
Expand each side carefully:
Left side: $6m^2 - 6m.$
Right side: $6m + m(m^2 - 3m + 2).$
Hence,
$$6m^2 - 6m = 6m + m^3 - 3m^2 + 2m.$$
Combine like terms:
$$6m^2 - 6m = m^3 + (6m + 2m) - 3m^2.$$
$$6m^2 - 6m = m^3 + 8m - 3m^2.$$
Step 6: Rearrange to Form a Quadratic in $m$
Bring all terms to one side:
$$6m^2 - 6m - \bigl(m^3 + 8m - 3m^2\bigr) = 0.$$
$$6m^2 - 6m - m^3 - 8m + 3m^2 = 0.$$
Combine like terms:
$$-m^3 + 9m^2 - 14m = 0.$$
Or equivalently:
$$m^3 - 9m^2 + 14m = 0.$$
Factoring out $m$ (if $m=0$ is trivial or not relevant for binomial expansions with positive $m$; we typically look for nonzero solutions):
$$m(m^2 - 9m + 14) = 0.$$
The quadratic part is
$$m^2 - 9m + 14 = 0.$$
Step 7: Match with the Given Options
The question provides four possible equations in $m$ and $r$. Substituting $r = 2$ into each choice and comparing which yields $m^2 - 9m + 14 = 0$ will identify the correct form.
Upon checking, the option that matches this condition is
$$(m^2) - m(4r + 1) + 4r^2 - 2 = 0.$$
For $r=2$, this becomes
$$m^2 - m(4 \cdot 2 + 1) + 4 \cdot (2)^2 - 2 = 0,$$
which simplifies to
$$m^2 - 9m + 14 = 0,$$
exactly matching our derived equation. Thus, the correct answer is
$$\boxed{(m^2) - m(4r + 1) + 4r^2 - 2 = 0.}$$
