When given two terms of an arithmetic sequence, you can find the common difference using the relationship between terms. Method 1 (Direct approach): The difference between the 7th and 3rd terms represents 4 common differences (from 3rd to 4th to 5th to 6th to 7th). Calculation: \(T_7 - T_3 = 4d\), so \(31 - 15 = 4d\), which gives \(16 = 4d\), therefore \(d = 4\). Method 2 (Using formula): \(T_3 = a + 2d = 15\) ... (equation 1) and \(T_7 = a + 6d = 31\) ... (equation 2). Subtract equation 1 from equation 2: \((a + 6d) - (a + 2d) = 31 - 15\), which simplifies to \(4d = 16\), so \(d = 4\). Both methods confirm d = 4. This type of problem is extremely common in ECZ Paper 1, where you must use algebraic reasoning. Understanding that the number of differences between terms equals the difference in their positions is key. Practice this skill as it's heavily tested.

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Zambia

• Dickson Tembo – 1 pts

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