QUESTION 4 [13] 4.1 Given the sequence \( 4 ; x ; 32 \). Determine the value(s) of the \( x \) if the sequence is: 4.1.1 Arithmetic. (2) 4.1.2 Geometric. (3) 4.2 Determine the value of \( P \) if: \[ P=\sum_{k=2}^{23} 3^{k-5} \] Tsthn Wharewinn Hethenal 4.3 Prove that for any arithmetic sequence of which the first term is \( a \) and the constant difference is \( d \), the sum to \( n \) terms can be expressed as \( S_{n}=\frac{n}{2}\left(2 a+\left(n^{\prime}-1\right) d\right) \). (4)

### 4.1 Sequence \(4,\, x,\, 32\) #### 4.1.1 Arithmetic Sequence In an arithmetic sequence, the middle term is the average of the first and third terms. Thus, we have \[ x = \frac{4 + 32}{2} = \frac{36}{2} = 18. \] #### 4.1.2 Geometric Sequence In a geometric sequence, the square of the middle term equals the product of the first and third terms: \[ x^2 = 4 \times 32 = 128. \] Taking the square root on both sides gives \[ x = \pm\sqrt{128} = \pm 8\sqrt{2}. \] --- ### 4.2 Evaluate \[ P=\sum_{k=2}^{23} 3^{k-5} \] **Step 1. Change the Index of Summation** Let \[ j=k-5. \] When \(k=2\), then \(j=2-5=-3\), and when \(k=23\), \(j=23-5=18\). Therefore, the sum becomes \[ P=\sum_{j=-3}^{18} 3^j. \] **Step 2. Apply the Geometric Series Formula** The sum of a geometric series with first term \(a\), common ratio \(r\) (with \(r\neq1\)), and \(n\) terms is given by \[ S_n=a\frac{r^n-1}{r-1}. \] Here, the first term is \(a=3^{-3}\), the common ratio is \(r=3\), and the number of terms is \[ n = 18 - (-3) + 1 = 22. \] Thus, \[ P = 3^{-3}\frac{3^{22} - 1}{3 - 1} = \frac{3^{-3}\left(3^{22} - 1\right)}{2}. \] Since \(3^{-3}=\frac{1}{27}\), \[ P = \frac{3^{22} - 1}{54}. \] --- ### 4.3 Proof of the Sum Formula for an Arithmetic Sequence Let an arithmetic sequence be given by \[ a,\, a+d,\, a+2d,\, \dots,\, a+(n-1)d. \] The sum of the first \(n\) terms is \[ S_n = a + (a+d) + (a+2d) + \cdots + \bigl(a+(n-1)d\bigr). \] **Step 1. Write the Sum in Two Ways** Write the sum in its original order: \[ S_n = a + (a+d) + (a+2d) + \cdots + \bigl(a+(n-1)d\bigr), \] and write it in reverse order: \[ S_n = \bigl(a+(n-1)d\bigr) + \bigl(a+(n-2)d\bigr) + \cdots + a. \] **Step 2. Add the Two Expressions Term by Term** Adding the two sums term by term gives: \[ 2S_n = \Bigl[a + \bigl(a+(n-1)d\bigr)\Bigr] + \Bigl[(a+d) + \bigl(a+(n-2)d\bigr)\Bigr] + \cdots. \] Notice that each pair sums to the same value: \[ a + \bigl(a+(n-1)d\bigr) = 2a + (n-1)d. \] Since there are \(n\) such pairs, we have \[ 2S_n = n\Bigl[2a + (n-1)d\Bigr]. \] **Step 3. Solve for \(S_n\)** Dividing both sides by 2: \[ S_n = \frac{n}{2}\Bigl[2a + (n-1)d\Bigr]. \] Thus, the formula for the sum of the first \(n\) terms of an arithmetic sequence is \[ S_n = \frac{n}{2}\Bigl[2a + (n-1)d\Bigr]. \]