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The first term of an AP is 3 and the sum of all 6 terms is 48. If the sum of all the terms is 168, calculate: a. the common difference, d b. the last term.

Mathematics

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Answers

2 parts
a
6156000
b
6076

Step-by-step solution

8 steps

Part a· 5 steps

Sum of all terms

  1. Sum formula for an AP

    Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} \left[2a + (n-1)d\right]

  2. Substitute n=2025n=2025, a=4a=4, d=3d=3

    S2025=20252[2(4)+(2025−1)(3)]S_{2025} = \frac{2025}{2} \left[2(4) + (2025-1)(3)\right] S2025=20252[8+2024×3]S_{2025} = \frac{2025}{2} \left[8 + 2024 \times 3\right]

  3. Simplify inside brackets

    2024×3=60722024 \times 3 = 6072 8+6072=60808 + 6072 = 6080 S2025=20252×6080S_{2025} = \frac{2025}{2} \times 6080

  4. Simplify

    60802=3040\frac{6080}{2} = 3040 S2025=2025×3040S_{2025} = 2025 \times 3040

  5. Compute 2025×30402025 \times 3040

    2000×3040=6 080 0002000 \times 3040 = 6\,080\,000 25×3040=76 00025 \times 3040 = 76\,000 6 080 000+76 000=6 156 0006\,080\,000 + 76\,000 = 6\,156\,000 6156000\boxed{6156000}

Part b6076

Last term

  1. Last term formula

    l=a+(n−1)dl = a + (n-1)d

  2. Substitute n=2025n=2025, a=4a=4, d=3d=3

    l=4+(2025−1)×3l = 4 + (2025-1) \times 3 l=4+2024×3l = 4 + 2024 \times 3

  3. Simplify

    2024×3=60722024 \times 3 = 6072 4+6072=60764 + 6072 = 6076 6076\boxed{6076}
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