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For the sequence 15, 12, 9,...., find the (a) tenth term, (b) nth term.

Mathematics

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2 parts
a
-12
b

18-3n

Step-by-step solution

7 steps

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The given sequence is 15,12,9,…15, 12, 9, \dots.
This is an arithmetic sequence.
The first term is a=15a = 15.
The common difference dd is found by subtracting any term from its succeeding term: d=12−15=−3d = 12 - 15 = -3.

Part a· 4 steps

To find the tenth term (T10T_{10}), we use the formula for the nn-th term of an arithmetic sequence

Tn=a+(n−1)dT_n = a + (n-1)d

  1. Substitute the values for aa, nn, and dd into the formula.

    Here, a=15a = 15, n=10n = 10, and d=−3d = -3.

    T10=15+(10−1)(−3)T_{10} = 15 + (10-1)(-3)

  2. Simplify the expression inside the parentheses.

    T10=15+(9)(−3)T_{10} = 15 + (9)(-3)

  3. Perform the multiplication.

    T10=15−27T_{10} = 15 - 27

  4. Perform the subtraction.

    T10=−12T_{10} = -12

    The tenth term is −12\boxed{-12}.

Part b18-3n

To find the nn-th term (TnT_n), we use the same formula and substitute the values for aa and dd.

Tn=a+(n−1)dT_n = a + (n-1)d

  1. Substitute a=15a = 15 and d=−3d = -3 into the formula.

    Tn=15+(n−1)(−3)T_n = 15 + (n-1)(-3)

  2. Distribute the −3-3 into the parentheses.

    Tn=15−3n+3T_n = 15 - 3n + 3

  3. Combine the constant terms.

    Tn=18−3nT_n = 18 - 3n

    The nn-th term is 18−3n\boxed{18-3n}.

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