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Let's solve **Question 11** step by step.

### 11a. **Find the value of \( a \):**

We are given the binomial expansion \( (4 + ax)^6 \) and told that one of the terms in this expansion is \( 160x^3 \). Our goal is to find the value of \( a \).

Using the binomial expansion formula for \( (4 + ax)^6 \):

\[
(4 + ax)^6 = \sum_{k=0}^{6} \binom{6}{k} (4)^{6-k} (ax)^k
\]

The general term in the expansion is:

\[
T_k = \binom{6}{k} (4)^{6-k} (ax)^k
\]

We are interested in the term where \( x^3 \) appears, so set \( k = 3 \) (since \( k \) corresponds to the power of \( x \)):

\[
T_3 = \binom{6}{3} (4)^{6-3} (ax)^3 = \binom{6}{3} (4)^3 a^3 x^3
\]

Now calculate \( \binom{6}{3} \) and \( 4^3 \):

\[
\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]
\[
4^3 = 64
\]

So, the term is:

\[
T_3 = 20 \times 64 \times a^3 x^3 = 1280a^3 x^3
\]

We are told that this term is equal to \( 160x^3 \). Therefore:

\[
1280a^3 = 160
\]

Solve for \( a^3 \):

\[
a^3 = \frac{160}{1280} = \frac{1}{8}
\]

Now, take the cube root of both sides:

\[
a = \frac{1}{2}
\]

So, the value of \( a \) is **\( \frac{1}{2} \)**.

---

### 11b. **Find the first two terms in the expansion of \( (4 + ax)^6 \) in ascending powers of \( x \):**

Now that we know \( a = \frac{1}{2} \), we can proceed with finding the first two terms in the binomial expansion of \( (4 + \frac{1}{2}x)^6 \).

The general binomial expansion formula is:

\[
(4 + \frac{1}{2}x)^6 = \sum_{k=0}^{6} \binom{6}{k} (4)^{6-k} \left( \frac{1}{2}x \right)^k
\]

We need the first two terms, which correspond to \( k = 0 \) and \( k = 1 \).

#### First term (\( k = 0 \)):

\[
T_0 = \binom{6}{0} (4)^6 \left( \frac{1}{2}x \right)^0 = 1 \times 4096 \times 1 = 4096
\]

#### Second term (\( k = 1 \)):

\[
T_1 = \binom{6}{1} (4)^5 \left( \frac{1}{2}x \right)^1 = 6 \times 1024 \times \frac{1}{2}x = 3072x
\]

### Final Answer:

- The first term is \( 4096 \).
- The second term is \( 3072x \).

---

Would you like further details or have any questions?

Here are five related questions for more practice:
1. What would the expansion be for \( (2 + bx)^5 \) for any \( b \)?
2. How can the binomial theorem be applied to negative exponents?
3. Can you derive the general term for the expansion of \( (x + y)^n \)?
4. How would the terms of \( (3 + 2x)^4 \) differ from this problem?
5. What would happen if \( a = -\frac{1}{2} \) in question 11a?

**Tip:** Always check the power of \( x \) carefully when identifying specific terms in a binomial expansion!

11. One of the terms in the binomial expansion of (4 + ax)^6 is 160x^3. 
a) Find the value of a. 
b) Using this value of a, find the first two terms in the expansion of (4 + ax)^6 in ascending powers of x.

Learn how to find the value of 'a' in the binomial expansion of (4 + ax)^6 where one of the terms is 160x^3, and how to expand the first two terms in ascending powers of x. This problem covers binomial theorem and combinatorial principles, suitable for A-Level Mathematics.

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