The first two non-zero terms in the expansion of $(1 + bx)(1 + ax)^6$ in ascending powers of $x$ are $1$ and $-\frac{21}{4}x^2$. Find the value of each of the constants $a$ and $b$, where $a < b$.

1. Expand $(1 + ax)^6$ up to the $x^2$ term.

$(1 + ax)^6 = 1 + \binom{6}{1}(ax) + \binom{6}{2}(ax)^2 + \dots$
$= 1 + 6ax + 15a^2x^2 + \dots$

2. Multiply the expansion by $(1 + bx)$ and collect terms up to $x^2$.

$(1 + bx)(1 + 6ax + 15a^2x^2 + \dots)$
$= 1 + 6ax + 15a^2x^2 + bx + 6abx^2 + \dots$
$= 1 + (6a + b)x + (15a^2 + 6ab)x^2 + \dots$

3. Equate the coefficient of $x$ to zero to find a relationship between $a$ and $b$.

Since the first two non-zero terms are $1$ and $-\frac{21}{4}x^2$, the coefficient of $x$ must be $0$.
$6a + b = 0$
$b = -6a$

4. Equate the coefficient of $x^2$ to $-\frac{21}{4}$ and solve for $a$.

$15a^2 + 6ab = -\frac{21}{4}$
Substitute $b = -6a$:
$15a^2 + 6a(-6a) = -\frac{21}{4}$
$15a^2 - 36a^2 = -\frac{21}{4}$
$-21a^2 = -\frac{21}{4}$
$a^2 = \frac{1}{4}$
$a = \frac{1}{2} \text{ or } a = -\frac{1}{2}$

5. Determine the correct values of $a$ and $b$ using the condition $a < b$.

If $a = \frac{1}{2}$, $b = -6(\frac{1}{2}) = -3$. Here $a > b$, which is rejected.
If $a = -\frac{1}{2}$, $b = -6(-\frac{1}{2}) = 3$. Here $a < b$, which is accepted.
Therefore, $a = -\frac{1}{2}$ and $b = 3$.