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For the geometric progression 189, 63, 21, ..., find: i) the common ratio ii) the 5th term iii) the sum to infinity.

Mathematics

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Answers

3 parts
i

13\frac{1}{3}

ii

73\frac{7}{3}

iii

5672\frac{567}{2}

Step-by-step solution

9 steps

Here's the solution for :

The given geometric progression (GP) is 189,63,21,…189, 63, 21, \dots
The first term is a=189a = 189.

Part i· 2 steps

Common ratio

  1. Calculate the common ratio rr by dividing a term by its preceding term.

    r=second termfirst term=63189r = \frac{\text{second term}}{\text{first term}} = \frac{63}{189}

  2. Simplify the fraction.

    r=63÷63189÷63=13r = \frac{63 \div 63}{189 \div 63} = \frac{1}{3}

    The common ratio is 13\boxed{\frac{1}{3}}.

Part ii

73\frac{7}{3}

5th term

  1. Use the formula for the nn-th term of a GP, Tn=arn−1T_n = ar^{n-1}.

    For the 5th term, n=5n=5.

    T5=189(13)5−1T_5 = 189 \left(\frac{1}{3}\right)^{5-1}

  2. Simplify the exponent and calculate the power.

    T5=189(13)4=189×1434=189×181T_5 = 189 \left(\frac{1}{3}\right)^4 = 189 \times \frac{1^4}{3^4} = 189 \times \frac{1}{81}

  3. Multiply and simplify the fraction.

    T5=18981=189÷2781÷27=73T_5 = \frac{189}{81} = \frac{189 \div 27}{81 \div 27} = \frac{7}{3}

    The 5th term is 73\boxed{\frac{7}{3}}.

Part iii

5672\frac{567}{2}

Sum to infinity

  1. Check if the sum to infinity exists.

    Since ∣r∣=∣13∣<1|r| = \left|\frac{1}{3}\right| < 1, the sum to infinity exists.

  2. Use the formula for the sum to infinity of a GP, S∞=a1−rS_\infty = \frac{a}{1-r}.

    S∞=1891−13S_\infty = \frac{189}{1 - \frac{1}{3}}

  3. Simplify the denominator.

    S∞=18933−13=18923S_\infty = \frac{189}{\frac{3}{3} - \frac{1}{3}} = \frac{189}{\frac{2}{3}}

  4. Perform the division.

    S∞=189×32=5672S_\infty = 189 \times \frac{3}{2} = \frac{567}{2}

    The sum to infinity is 5672\boxed{\frac{567}{2}}.

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