UESTION 7 1. Consider the geometric sequence: \( 4 ;-2 ; 1 \ldots \) 1... 1.2 Determine the next term of the sequence. 1.3 Calculate the sum to infinity of the series \( 4-2+1 \ldots \) 1... If \( x \) is a real number, show that the following sequence can NOT be geometric: \( 1 ; x+1: x-3 \ldots \)

1. **Next term in the geometric sequence \(4, -2, 1, \ldots\)** The common ratio is given by \[ r=\frac{-2}{4}=-\frac{1}{2}. \] The third term is \(1\), so the fourth term is \[ 1 \times \left(-\frac{1}{2}\right)=-\frac{1}{2}. \] 2. **Sum to infinity of the series \(4-2+1+\ldots\)** For a geometric series with first term \(a\) and common ratio \(r\) (with \(|r|<1\)), the sum to infinity is given by \[ S_{\infty}=\frac{a}{1-r}. \] Here, \(a=4\) and \(r=-\frac{1}{2}\). Thus, \[ S_{\infty}=\frac{4}{1-\left(-\frac{1}{2}\right)}=\frac{4}{1+\frac{1}{2}}=\frac{4}{\frac{3}{2}}=\frac{8}{3}. \] 3. **Proof that the sequence \(1,\; x+1,\; x-3,\ldots\) cannot be geometric if \(x\) is real** For a sequence to be geometric, there exists a common ratio \(r\) such that \[ x+1 = 1 \cdot r \quad \text{and} \quad x-3 = (x+1) \cdot r. \] From the first relation, the common ratio is \[ r=x+1. \] Substituting into the second relation gives: \[ x-3=(x+1)^2. \] Expanding and rearranging, \[ x-3=x^2+2x+1 \quad \Longrightarrow \quad 0=x^2+2x+1 - x+3, \] \[ x^2 + x + 4 = 0. \] Calculating the discriminant, \[ \Delta = 1^2 - 4\cdot1\cdot4 = 1-16=-15. \] Since the discriminant is negative, the quadratic equation has no real solutions. Hence, there is no real number \(x\) for which the sequence is geometric.