QUESTION 3 The first three terms of au arithmetic sequence are $$ 2 p-3 ; p+5 ; 2 p+7 $$. 3.1 Deternine the ralue(s) of $$ p $$. (3) 3.2 Calculate the sum of the first 120 terms. (3) 3.3 The following pattern is true for the arithmetic sequence above: $$ \begin{array}{l} T_{1}+T_{4}=T_{2}+T_{3} \ T_{5}+T_{8}=T_{6}+T_{7} \ T_{9}+T_{12}=T_{10}+T_{11} \ \therefore T_{k}+T_{k+3}=T_{x}+T_{y} \end{array} $$ 3.3.1 Write down the values of $$ x $$ and $$ y $$ in terms of $$ k $$. (2) 3.32 Hence. calculate the ralue of $$ T_{x}+T_{y} $$ in terms of $$ k $$ in simplest form (4)

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Let's tackle each part of **Question 3** step by step.

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### **3.1 Determine the value(s) of $$ p $$.**

Given the first three terms of an arithmetic sequence:
- $$ T_1 = 2p - 3 $$
- $$ T_2 = p + 5 $$
- $$ T_3 = 2p + 7 $$

In an arithmetic sequence, the difference between consecutive terms is constant. Therefore:

$$
T_2 - T_1 = T_3 - T_2
$$

Substituting the given terms:

$$
(p + 5) - (2p - 3) = (2p + 7) - (p + 5)
$$

Simplify both sides:

$$
-p + 8 = p + 2
$$

Combine like terms:

$$
-2p = -6 \quad \Rightarrow \quad p = 3
$$

**Answer:** $$ p = 3 $$

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### **3.2 Calculate the sum of the first 120 terms.**

First, determine the common difference $$ d $$ and the first term $$ a_1 $$ using $$ p = 3 $$:

$$
a_1 = 2p - 3 = 2(3) - 3 = 3
$$
$$
a_2 = p + 5 = 3 + 5 = 8 \quad \Rightarrow \quad d = a_2 - a_1 = 5
$$

The sum of the first $$ n $$ terms of an arithmetic sequence is given by:

$$
S_n = \frac{n}{2} [2a_1 + (n - 1)d]
$$

Substituting $$ n = 120 $$, $$ a_1 = 3 $$, and $$ d = 5 $$:

$$
S_{120} = \frac{120}{2} [2(3) + (120 - 1)(5)] = 60 [6 + 595] = 60 \times 601 = 36,\!060
$$

**Answer:** The sum of the first 120 terms is **36,060**.

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### **3.3 The following pattern is true for the arithmetic sequence above:**

$$
\begin{array}{l}
T_{1}+T_{4}=T_{2}+T_{3} \
T_{5}+T_{8}=T_{6}+T_{7} \
T_{9}+T_{12}=T_{10}+T_{11} \
\vdots \
\therefore T_{k}+T_{k+3}=T_{x}+T_{y}
\end{array}
$$

#### **3.3.1 Write down the values of $$ x $$ and $$ y $$ in terms of $$ k $$.**

Observing the pattern:

- When $$ k = 1 $$: $$ x = 2 $$, $$ y = 3 $$
- When $$ k = 5 $$: $$ x = 6 $$, $$ y = 7 $$
- When $$ k = 9 $$: $$ x = 10 $$, $$ y = 11 $$
- ...

It becomes clear that:

$$
x = k + 1 \quad \text{and} \quad y = k + 2
$$

**Answer:** 
$$
x = k + 1 \quad \text{and} \quad y = k + 2
$$

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#### **3.3.2 Calculate the value of $$ T_{x} + T_{y} $$ in terms of $$ k $$ in simplest form.**

Given $$ x = k + 1 $$ and $$ y = k + 2 $$, and knowing that in an arithmetic sequence:

$$
T_n = a_1 + (n - 1)d
$$

Compute $$ T_x $$ and $$ T_y $$:

$$
T_x = a_1 + (k + 1 - 1)d = a_1 + k d
$$
$$
T_y = a_1 + (k + 2 - 1)d = a_1 + (k + 1)d
$$

Add them together:

$$
T_x + T_y = (a_1 + k d) + (a_1 + (k + 1)d) = 2a_1 + (2k + 1)d
$$

Substitute $$ a_1 = 3 $$ and $$ d = 5 $$:

$$
T_x + T_y = 2(3) + (2k + 1)(5) = 6 + 10k + 5 = 10k + 11
$$

**Answer:** 
$$
T_{x} + T_{y} = 10k + 11
$$

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**3.1** $$ p = 3 $$ **3.2** Sum of the first 120 terms is **36,060**. **3.3.1** $$ x = k + 1 $$ $$ y = k + 2 $$ **3.3.2** $$ T_{x} + T_{y} = 10k + 11 $$

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