A teacher's initial pay is K850 per fortnight, increasing by K120 per fortnight each year. Calculate his pay for the first 5 years, in Year 10, and when it first exceeds K1500.

You're on a roll — let's solve this problem about the teacher's pay. The teacher's initial pay is K850 per fortnight, and it increases by K120 per fortnight each year. This forms an arithmetic sequence where: First term, a = 850 Common difference, d = 120 The formula for the n-th term of an arithmetic sequence is T_n = a + (n-1)d. a) Write his fortnightly pay as an arithmetic sequence for the first 5 years. Step 1: Calculate the pay for each of the first five years using T_n = a + (n-1)d. For Year 1 (n=1): T_1 = 850 + (1-1)120 = 850 + 0 = 850 For Year 2 (n=2): T_2 = 850 + (2-1)120 = 850 + 120 = 970 For Year 3 (n=3): T_3 = 850 + (3-1)120 = 850 + 2(120) = 850 + 240 = 1090 For Year 4 (n=4): T_4 = 850 + (4-1)120 = 850 + 3(120) = 850 + 360 = 1210 For Year 5 (n=5): T_5 = 850 + (5-1)120 = 850 + 4(120) = 850 + 480 = 1330 Answer: K850, K970, K1090, K1210, K1330 b) What will his fortnightly pay be in Year 10? Step 1: Use the formula T_n = a + (n-1)d with n=10. T_10 = 850 + (10-1)120 Step 2: Calculate the value. T_10 = 850 + (9)120 T_10 = 850 + 1080 T_10 = 1930 Answer: K1930 c) In which year will his fortnightly pay first exceed K1 500? Step 1: Set up an inequality using the formula T_n = a + (n-1)d and solve for n. 850 + (n-1)120 > 1500 Step 2: Isolate the term with n. (n-1)120 > 1500 - 850 (n-1)120 > 650 Step 3: Divide by 120. n-1 > (650)/(120) n-1 > 5.416... Step 4: Solve for n. n > 5.416... + 1 n > 6.416... Step 5: Since n must be an integer (representing a year), the smallest integer greater than 6.416... is 7. Answer: Year 7 What's next?