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Anonymous asked in Science & MathematicsMathematics · 1 month ago

Find the 5th term of the arithmetic sequence whose 3rd term is 35 and whose 10th term is 77?

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  • 1 month ago

    The 5th term is 47.

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  • 1 month ago

    T(3)=35=>T(1)+(3-1)d=35---------(1)

    T(10)=77=>T(1)+(10-1)d=77-------(2)

    (2)-(1)=>

    7d=42=>d=6 (the common difference)

    T(1)=35-2(6)=23

    Thus

    T(5)=23+(5-1)6=47

  • 1 month ago

    The 5th term of the arithmetic sequence 

    whose 3rd term is 35 

    and whose 10th term is 77:

    47

  • 1 month ago

    For an arithmetic sequence:

    a₁

    a₂ = a₁ + d ← where d is the common difference

    a₃ = a₂ + d = a₁ + 2d

    a₄ = a₃ + d = a₁ + 3d

    a₅ = a₄ + d = a₁ + 4d

    …and you can generalize by writing

    a(n) = a₁ + (n - 1).d

    a(n) = a₁ + (n - 1).d → when: n = 3

    a₃ = a₁ + 2d → given that it's 35

    a₁ + 2d = 35

    a₁ = 35 - 2d

    a(n) = a₁ + (n - 1).d → when: n = 10

    a₁₀ = a₁ + 9d → given that it's 77

    a₁ + 9d = 77 → recall the previous result: a₁ = 35 - 2d

    35 - 2d + 9d = 77

    7d = 42

    d = 6

    Recall:

    a₁ = 35 - 2d → we've just seen that: d = 6

    a₁ = 35 - 12

    a₁ = 23

    a₅ = a₁ + 4d

    a₅ = 23 + (4 * 6)

    a₅ = 47

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  • t[3] = 35 = m + (3 - 1) * d = m + 2d

    t[10] = 77 = m + (10 - 1) * d = m + 9d

    77 - 35 = m + 9d - (m + 2d)

    42 = m - m + 9d - 2d

    42 = 7d

    6 = d

    35 = m + 2d

    35 = m + 2 * 6

    35 = m + 12

    23 = m

    t[n] = m + (n - 1) * d

    t[n] = 23 + (n - 1) * 6

    t[n] = 23 + 6n - 6

    t[n] = 17 + 6n

    t[5] = 17 + 6 * 5 = 17 + 30 = 47

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