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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.

Mathematics

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Answers

3 parts
a

K850, K970, K1090, K1210, K1330

b

K1930

c

Year 7\text{Year 7}

Step-by-step solution

8 steps

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=850a = 850
Common difference, d=120d = 120
The formula for the nn-th term of an arithmetic sequence is Tn=a+(n−1)dT_n = a + (n-1)d.

Part a· 1 step

Write his fortnightly pay as an arithmetic sequence for the first 5 years.

  1. Calculate the pay for each of the first five years using Tn=a+(n−1)dT_n = a + (n-1)d.

    For Year 1 (n=1n=1): T1=850+(1−1)120=850+0=850T_1 = 850 + (1-1)120 = 850 + 0 = 850
    For Year 2 (n=2n=2): T2=850+(2−1)120=850+120=970T_2 = 850 + (2-1)120 = 850 + 120 = 970
    For Year 3 (n=3n=3): T3=850+(3−1)120=850+2(120)=850+240=1090T_3 = 850 + (3-1)120 = 850 + 2(120) = 850 + 240 = 1090
    For Year 4 (n=4n=4): T4=850+(4−1)120=850+3(120)=850+360=1210T_4 = 850 + (4-1)120 = 850 + 3(120) = 850 + 360 = 1210
    For Year 5 (n=5n=5): T5=850+(5−1)120=850+4(120)=850+480=1330T_5 = 850 + (5-1)120 = 850 + 4(120) = 850 + 480 = 1330
    Answer: K850,K970,K1090,K1210,K1330\boxed{K850, K970, K1090, K1210, K1330}

Part bK1930

What will his fortnightly pay be in Year 10?

  1. Use the formula Tn=a+(n−1)dT_n = a + (n-1)d with n=10n=10.

    T10=850+(10−1)120T_{10} = 850 + (10-1)120
  2. Calculate the value.

    T10=850+(9)120T_{10} = 850 + (9)120 T10=850+1080T_{10} = 850 + 1080 T10=1930T_{10} = 1930

    Answer: K1930\boxed{K1930}

Part c

Year 7\text{Year 7}

In which year will his fortnightly pay first exceed K1 500?

  1. Set up an inequality using the formula Tn=a+(n−1)dT_n = a + (n-1)d and solve for nn.

    850+(n−1)120>1500850 + (n-1)120 > 1500
  2. Isolate the term with nn.

    (n−1)120>1500−850(n-1)120 > 1500 - 850 (n−1)120>650(n-1)120 > 650
  3. Divide by 120.

    n−1>650120n-1 > \frac{650}{120} n−1>5.416...n-1 > 5.416...
  4. Solve for nn.

    n>5.416...+1n > 5.416... + 1 n>6.416...n > 6.416...
  5. Since nn must be an integer (representing a year), the smallest integer greater than 6.416...

    is 7.
    Answer: Year 7\boxed{\text{Year 7}}

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