QUESTION 3 $$ 3.1 \quad $$ The first three terms of an arithmetic sequence are: $$ 2 p+3 ; p+6 $$ and $$ p-2 $$. $$ \begin{array}{lll}3.1 .1 & ext { Show that } p=11 \ 3.1 .2 & ext { Calculate the smallest value of } n ext { for which } T_{n}

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QUESTION 3 $$ 3.1 \quad $$ The first three terms of an arithmetic sequence are: $$ 2 p+3 ; p+6 $$ and $$ p-2 $$. $$ \begin{array}{lll}3.1 .1 & 	ext { Show that } p=11 \ 3.1 .2 & 	ext { Calculate the smallest value of } n 	ext { for which } T_{n}<-55\end{array} $$ 3.2 $$ \quad $$ Given that $$ \sum_{k=1}^{6}(x-3 k)=\sum_{k=1}^{9}(x-3 k) $$, prove that $$ \sum_{k=1}^{15}(x-3 k)=0 $$.

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**3.1.1**

The first three terms of the arithmetic sequence are  
$$
T_1 = 2p+3,\quad T_2 = p+6,\quad T_3 = p-2.
$$

In an arithmetic sequence, the difference between consecutive terms is constant. Therefore,  
$$
T_2-T_1 = T_3-T_2.
$$

Substitute the terms:
$$
(p+6) - (2p+3) = (p-2) - (p+6).
$$

Simplify each side:
$$
p + 6 - 2p - 3 = p - 2 - p - 6,
$$
$$
-p + 3 = -8.
$$

Solve for $$ p $$:
$$
-p = -11 \quad \Longrightarrow \quad p = 11.
$$

---

**3.1.2**

Now, substitute $$ p=11 $$ into the terms to find the first term and the common difference:
$$
T_1 = 2(11)+3 = 22+3 = 25,
$$
$$
T_2 = 11+6 = 17,
$$
$$
T_3 = 11-2 = 9.
$$

The common difference $$ d $$ is:
$$
d = T_2 - T_1 = 17 - 25 = -8.
$$

The general term of an arithmetic sequence is given by:
$$
T_n = T_1 + (n-1)d.
$$

Substitute $$ T_1 = 25 $$ and $$ d = -8 $$:
$$
T_n = 25 + (n-1)(-8) = 25 - 8(n-1) = 25 - 8n + 8 = 33 - 8n.
$$

We need to find the smallest integer $$ n $$ such that:
$$
T_n < -55.
$$

Substitute $$ T_n = 33 - 8n $$:
$$
33 - 8n < -55.
$$

Solve this inequality:
1. Subtract 33 from both sides:
   $$
   -8n < -88.
   $$
2. Divide both sides by $$-8$$ (remember to reverse the inequality sign when dividing by a negative number):
   $$
   n > 11.
   $$

The smallest integer greater than 11 is:
$$
n = 12.
$$

---

**3.2**

We are given:
$$
\sum_{k=1}^{6}(x-3k)=\sum_{k=1}^{9}(x-3k).
$$

First, compute each sum.

For the sum from $$ k=1 $$ to 6:
$$
\sum_{k=1}^{6}(x-3k) = 6x - 3\sum_{k=1}^{6} k.
$$
The sum of the first 6 natural numbers is:
$$
\sum_{k=1}^{6} k = \frac{6 \cdot 7}{2} = 21.
$$
So,
$$
\sum_{k=1}^{6}(x-3k) = 6x - 3 \cdot 21 = 6x - 63.
$$

For the sum from $$ k=1 $$ to 9:
$$
\sum_{k=1}^{9}(x-3k) = 9x - 3\sum_{k=1}^{9} k.
$$
The sum of the first 9 natural numbers is:
$$
\sum_{k=1}^{9} k = \frac{9 \cdot 10}{2} = 45.
$$
Thus,
$$
\sum_{k=1}^{9}(x-3k) = 9x - 3 \cdot 45 = 9x - 135.
$$

Equate the two sums:
$$
6x - 63 = 9x - 135.
$$

Solve for $$ x $$:
1. Rearranging:
   $$
   9x - 135 - 6x + 63 = 0 \quad \Longrightarrow \quad 3x - 72 = 0.
   $$
2. Hence, 
   $$
   3x = 72 \quad \Longrightarrow \quad x = 24.
   $$

Now, we need to show that:
$$
\sum_{k=1}^{15}(x-3k)=0.
$$

Substitute $$ x = 24 $$:
$$
\sum_{k=1}^{15}(24-3k) = 15 \cdot 24 - 3\sum_{k=1}^{15} k.
$$

Calculate the sum of the first 15 natural numbers:
$$
\sum_{k=1}^{15} k = \frac{15 \cdot 16}{2} = 120.
$$

Thus,
$$
\sum_{k=1}^{15}(24-3k) = 360 - 3 \cdot 120 = 360 - 360 = 0.
$$

This completes the proof.
**3.1.1** Show that $$ p = 11 $$. **3.1.2** The smallest value of $$ n $$ for which $$ T_n < -55 $$ is $$ n = 12 $$. **3.2** Given that $$ \sum_{k=1}^{6}(x-3k) = \sum_{k=1}^{9}(x-3k) $$, it follows that $$ \sum_{k=1}^{15}(x-3k) = 0 $$.

QUESTION 3
The first three terms of an arithmetic sequence are: and .

3.2 Given that , prove that .

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QUESTION 3 The first three terms of an arithmetic sequence are: and . 3.2 Given that , prove that .