Q.If the coefficients of $2^{\text{nd}}$, $3^{\text{rd}}$ and the $4^{\text{th}}$ terms in the expansion of $(1 + x)^n$ are in A.P., then value of $n$ is
(A) $2$
(B) $7$
(C) $11$
(D) $14$

The key idea involves identifying the coefficients of the binomial expansion and applying the condition for an Arithmetic Progression (A.P.).

1. The general term in the expansion of $(1+x)^n$ is $T_{r+1} = \binom{n}{r} x^r$.

   • The coefficient of the $2^{\text{nd}}$ term ($T_2$) is $\binom{n}{1}$.
   • The coefficient of the $3^{\text{rd}}$ term ($T_3$) is $\binom{n}{2}$.
   • The coefficient of the $4^{\text{th}}$ term ($T_4$) is $\binom{n}{3}$.

2. Since these coefficients are in A.P., we have:

   $2 \binom{n}{2} = \binom{n}{1} + \binom{n}{3}$

3. Substitute the formulas for binomial coefficients:

   $2 \left( \frac{n(n-1)}{2} \right) = n + \frac{n(n-1)(n-2)}{3 \times 2 \times 1}$

   $n(n-1) = n + \frac{n(n-1)(n-2)}{6}$

4. For the $4^{\text{th}}$ term to exist and have a non-zero coefficient, $n$ must be at least $3$. Thus, we can divide by $n$ (since $n \ne 0$):

   $n-1 = 1 + \frac{(n-1)(n-2)}{6}$ …

When we expand a binomial expression like $(1+x)^n$, each term has a coefficient. The problem states that three specific coefficients are in an arithmetic progression (A.P.). Our task is to use this condition to find the value of $n$.

The core idea is to first identify the general form of a term in the binomial expansion and then extract the coefficients for the specified terms. Once we have these coefficients, we apply the property of an A.P. to set up an equation and solve for $n$.

1. Identify the coefficients of the specified terms.

   • The $2^{\text{nd}}$ term corresponds to $r+1=2$, so $r=1$. Its coefficient is $\binom{n}{1}$.
   • The $3^{\text{rd}}$ term corresponds to $r+1=3$, so $r=2$. Its coefficient is $\binom{n}{2}$.
   • The $4^{\text{th}}$ term corresponds to $r+1=4$, so $r=3$. Its coefficient is $\binom{n}{3}$.

2. Apply the A.P. condition.

   If three numbers $a, b, c$ are in A.P., then the middle term is the average of the other two, which means $2b = a+c$.

   Here, our coefficients are $\binom{n}{1}$, $\binom{n}{2}$, and $\binom{n}{3}$.

   So, we have the equation:

3. Expand the binomial coefficients.

   Recall the definition of binomial coefficients: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

   • $\binom{n}{1} = \frac{n!}{1!(n-1)!} = n$
   • $\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2}$
   • $\binom{n}{3} = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)}{6}$

4. Substitute these expressions into the A.P. equation.

Simplify the left side:

5. Solve the resulting equation for $n$. For the $4^{\text{th}}$ term to exist, $n$ must be at least $3$. This means $n \neq 0$. We can divide the entire equation by $n$:

Multiply the entire equation by 6 to clear the denominator:

Expand both sides: …