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Find the sum of the first 10 terms of the following sequences: a) 8, 3, -2, -7, ... b) 3, 1, 1/3, 1/9, ...

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2 parts
a
-145
b

295246561\frac{29524}{6561}

Step-by-step solution

5 steps

4) Find the sum of the first 10 terms of the following sequences

Part a· 2 steps

8, 3, -2, -7, ...

  1. Identify the type of sequence and its parameters.

    As determined in question 3a, this is an arithmetic sequence with the first term a1=8a_1 = 8 and common difference d=−5d = -5. We need to find the sum of the first 10 terms, so n=10n=10.

  2. Use the formula for the sum of the first nn terms of an arithmetic sequence.

    The formula is Sn=n2(2a1+(n−1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d).
    Substitute n=10n=10, a1=8a_1=8, and d=−5d=-5:

    S10=102(2(8)+(10−1)(−5))S_{10} = \frac{10}{2}(2(8) + (10-1)(-5)) S10=5(16+(9)(−5))S_{10} = 5(16 + (9)(-5)) S10=5(16−45)S_{10} = 5(16 - 45) S10=5(−29)S_{10} = 5(-29) S10=−145S_{10} = -145

    The sum of the first 10 terms is −145\boxed{-145}.

Part b

295246561\frac{29524}{6561}

3, 1, 13\frac{1}{3}, 19\frac{1}{9}, ...

  1. Identify the type of sequence and its parameters.

    As determined in question 3b, this is a geometric sequence with the first term a1=3a_1 = 3 and common ratio r=13r = \frac{1}{3}. We need to find the sum of the first 10 terms, so n=10n=10.

  2. Use the formula for the sum of the first nn terms of a geometric sequence.

    The formula is Sn=a1(1−rn)1−rS_n = \frac{a_1(1-r^n)}{1-r}.
    Substitute n=10n=10, a1=3a_1=3, and r=13r=\frac{1}{3}:

    S10=3(1−(13)10)1−13S_{10} = \frac{3\left(1 - \left(\frac{1}{3}\right)^{10}\right)}{1 - \frac{1}{3}}

  3. Simplify the expression.

    S10=3(1−1310)23S_{10} = \frac{3\left(1 - \frac{1}{3^{10}}\right)}{\frac{2}{3}} S10=3(310−1310)23S_{10} = \frac{3\left(\frac{3^{10}-1}{3^{10}}\right)}{\frac{2}{3}} S10=3⋅310−1310⋅32S_{10} = 3 \cdot \frac{3^{10}-1}{3^{10}} \cdot \frac{3}{2} S10=32(310−1)2⋅310S_{10} = \frac{3^2(3^{10}-1)}{2 \cdot 3^{10}} S10=9(310−1)2⋅310S_{10} = \frac{9(3^{10}-1)}{2 \cdot 3^{10}} S10=9(59049−1)2⋅59049S_{10} = \frac{9(59049-1)}{2 \cdot 59049} S10=9(59048)118098S_{10} = \frac{9(59048)}{118098} S10=531432118098S_{10} = \frac{531432}{118098} S10=26571659049S_{10} = \frac{265716}{59049}

    This fraction can be simplified by dividing the numerator and denominator by 9:

    S10=5904813122S_{10} = \frac{59048}{13122}

    Further simplification by dividing by 2:

    S10=295246561S_{10} = \frac{29524}{6561}

    The sum of the first 10 terms is 295246561\boxed{\frac{29524}{6561}}.

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