QUESTION 7 [12) 7.1 Consider the geometric sequence: \( 4 ;-2 ; 1 \)... 7.1.1 Determine the next term of the sequence. 7.1.2 Determine \( n \) if the \( n^{\text {th }} \) term is \( \frac{1}{64} \ \). (2) 7.1.3 Calculate the sum to infinity of the series \( 4-2+1 \ldots \) (4) 7.2 If \( x \) is a real number, show that the following sequence can NOT be geometric: \( 1 ; x+1 ; x-3 \ldots \) (2)

**7.1.1** The first term is \( a = 4 \) and the second term is \(-2\). Thus, the common ratio is \[ r=\frac{-2}{4}=-\frac{1}{2}. \] The next term after \(1\) is obtained by multiplying by \( r \): \[ 1 \cdot \left(-\frac{1}{2}\right)=-\frac{1}{2}. \] **7.1.2** The general term of a geometric sequence is given by \[ a_n = a\, r^{n-1}. \] Here, \( a = 4 \) and \( r = -\frac{1}{2} \). We are given \[ 4\left(-\frac{1}{2}\right)^{n-1}=\frac{1}{64}. \] Divide both sides by 4: \[ \left(-\frac{1}{2}\right)^{n-1}=\frac{1}{256}. \] Since \(\frac{1}{256}\) is positive, the exponent \( n-1 \) must be even (because a negative base raised to an even power is positive). Let \( n-1=2k \), then \[ \left(-\frac{1}{2}\right)^{2k}=\left(\frac{1}{4}\right)^k. \] So, \[ \left(\frac{1}{4}\right)^k=\frac{1}{256}. \] Recognize that \[ 256=4^4 \quad \text{so} \quad \frac{1}{256}=\frac{1}{4^4}. \] Thus, \[ \left(\frac{1}{4}\right)^k=\left(\frac{1}{4}\right)^4. \] Hence, \[ k=4 \quad \text{and} \quad n-1=8. \] Therefore, \[ n=9. \] **7.1.3** The sum to infinity for a geometric series with \(|r|<1\) is given by \[ S_{\infty}=\frac{a}{1-r}. \] Here, \( a = 4 \) and \( r=-\frac{1}{2} \). Hence, \[ S_{\infty}=\frac{4}{1-(-\frac{1}{2})}=\frac{4}{1+\frac{1}{2}}=\frac{4}{\frac{3}{2}}=\frac{8}{3}. \] **7.2** Consider the sequence: \[ 1,\quad x+1,\quad x-3,\ldots \] For a geometric sequence, the ratio between consecutive terms must be constant. The ratio from the first to the second term is: \[ r=x+1. \] The ratio from the second to the third term is: \[ r=\frac{x-3}{x+1} \quad \text{(provided } x+1\neq 0\text{)}. \] For the sequence to be geometric we require: \[ x+1=\frac{x-3}{x+1}. \] Multiply both sides by \( x+1 \) (assuming \( x\neq -1 \)): \[ (x+1)^2=x-3. \] This expands to: \[ x^2+2x+1=x-3. \] Subtract \( x-3 \) from both sides: \[ x^2+2x+1-x+3=0 \quad \Longrightarrow \quad x^2+x+4=0. \] The discriminant of this quadratic is: \[ \Delta=1^2-4(1)(4)=1-16=-15. \] Since \( \Delta<0 \), there are no real solutions for \( x \). If \( x=-1 \), then the second term becomes \( x+1=0 \) and the ratio from the first to the second term is 0, but the third term \( x-3=-4 \) does not satisfy a constant ratio. Thus, there is no real value of \( x \) that makes the sequence geometric.