MATHEMATICS DEPARTMENT HOLHDAY ASSIGNMENT GRADE 11 Due date: first week of reporting \[ \left.s(-1)^{-1}\right)-2 \] 1. The first 3 terms of a geometric progression are \( 3 a-2,3 a+6 \) and \( 3 a+30 \). Find the 5. \( T_{1}=\frac{T_{1}}{T_{2}}=\frac{T_{2}}{T_{3}} \) a) Value of a b) \( \mathbf{N}^{\text {th }} \) term c) sum of the first 4 terms \[ \frac{3 a-2}{3 a+6} \times \frac{3 a+6}{3 a+30} \]

To solve the problem, let’s dive into the world of geometric progressions (GPs)! Geometric progressions are sequences where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In your case, the first three terms are \(3a - 2\), \(3a + 6\), and \(3a + 30\). Setting the ratio of successive terms equal gives us a neat equation; don't forget to simplify! Now, for the \(N^{\text{th}}\) term of a GP, the formula is given by \(T_n = T_1 \cdot r^{(n-1)}\), where \(r\) is the common ratio you would find from the terms. And for summing the first four terms, the sum \(S_n\) of the first \(n\) terms is given by \(S_n = T_1 \cdot \frac{1 - r^n}{1 - r}\) (as long as \(r \neq 1\)). Just plug in your values for \(T_1\) and \(r\) once you find them! A common mistake here is mixing up the order of operations. Remember to keep your ratios consistent. Not simplifying correctly or missing negative signs can trip you up when calculating the common ratio. So, double-check your math and keep an eye on those divine little digits!