Let us consider the binomial expansion ` (1 + x)^(n) = sum_(r=0)^(n) a_(r) x^(r)`
where ` a_(4) , a_(5) "and " a_(6) ` are in AP , ( n ` lt ` 10 ). Consider another
binomial expansion of ` A = root (3)(2) + (root(4) (3))^(13n)` , the expansion of A
contains some rational terms ` T_(a1),T_(a2),T_(a3),...,T_(am)`
` (a_(1) lt a_(2) lt a_(3) lt ...lt a_(m))`
The value of ` a_(m)` is

Options

A. 87
B. 88
C. 89
D. 90

Explanation

To solve the given problem step by step, we will break it down into manageable parts.

### Step 1: Understanding the Binomial Expansion Condition
We start with the binomial expansion of \( (1 + x)^n \) where the coefficients \( a_4, a_5, a_6 \) are in Arithmetic Progression (AP). The coefficients are given by:
\[ a_r = \binom{n}{r} \]
Thus, we have:
- \( a_4 = \binom{n}{4} \)
- \( a_5 = \binom{n}{5} \)
- \( a_6 = \binom{n}{6} \)

For these to be in AP, the condition is:
\[ 2a_5 = a_4 + a_6 \]

### Step 2: Setting Up the Equation
Substituting the binomial coefficients into the AP condition, we get:
\[ 2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \]

Using the properties of binomial coefficients, we can express \( \binom{n}{4} \) and \( \binom{n}{6} \) in terms of \( \binom{n}{5} \):
\[ \binom{n}{4} = \frac{n-4}{5} \binom{n}{5} \]
\[ \binom{n}{6} = \frac{6}{n-5} \binom{n}{5} \]

### Step 3: Simplifying the Equation
Substituting these into the AP condition gives:
\[ 2 \binom{n}{5} = \frac{n-4}{5} \binom{n}{5} + \frac{6}{n-5} \binom{n}{5} \]
Dividing through by \( \binom{n}{5} \) (assuming \( n \geq 6 \)):
\[ 2 = \frac{n-4}{5} + \frac{6}{n-5} \]

### Step 4: Finding a Common Denominator
Multiplying through by \( 5(n-5) \) to eliminate the denominators:
\[ 10(n-5) = (n-4)(n-5) + 30 \]
Expanding both sides:
\[ 10n - 50 = n^2 - 9n + 20 \]
Rearranging gives:
\[ n^2 - 19n + 70 = 0 \]

### Step 5: Solving the Quadratic Equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{19 \pm \sqrt{(-19)^2 - 4 \cdot 1 \cdot 70}}{2 \cdot 1} = \frac{19 \pm \sqrt{361 - 280}}{2} = \frac{19 \pm \sqrt{81}}{2} = \frac{19 \pm 9}{2} \]
This gives:
\[ n = 14 \quad \text{or} \quad n = 5 \]
Since \( n < 10 \), we take \( n = 7 \).

### Step 6: Analyzing the Second Binomial Expansion
Now consider the second binomial expansion:
\[ A = \left(2^{1/3} + 3^{1/4}\right)^{13n} = \left(2^{1/3} + 3^{1/4}\right)^{91} \]
The general term \( T_r \) in this expansion is given by:
\[ T_r = \binom{91}{r} (2^{1/3})^{91-r} (3^{1/4})^r \]

### Step 7: Finding Rational Terms
For \( T_r \) to be rational, the exponent of \( 3 \) must be an integer, which means:
\[ \frac{r}{4} \text{ must be an integer} \implies r = 0, 4, 8, \ldots, 88 \]
The maximum value of \( r \) that satisfies this condition and is less than or equal to 91 is \( 88 \).

### Final Answer
Thus, the value of \( a_m \) is:
\[ \boxed{88} \]