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In a geometric sequence, the sum of the first two terms is 135, and the sum of the third and fourth terms is 60. If the common ratio is positive, calculate the common ratio, the first term, and the su

Mathematics

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Answers

2 parts
i

23\frac{2}{3}

81
ii
243

Step-by-step solution

4 steps

Let the first term of the exponential sequence be aa and the common ratio be rr.
The terms of a geometric sequence are a,ar,ar2,ar3,…a, ar, ar^2, ar^3, \dots.

Given information:

  1. The sum of the first two terms is 135: a+ar=135a + ar = 135 a(1+r)=135a(1+r) = 135 (Equation 1)

  2. The sum of the third and fourth terms is 60: ar2+ar3=60ar^2 + ar^3 = 60 ar2(1+r)=60ar^2(1+r) = 60 (Equation 2)

  3. The common ratio rr is positive.

Part i· 2 steps

Calculate the common ratio and the first term.

  1. Divide Equation 2 by Equation 1 to find the common ratio rr.

    ar2(1+r)a(1+r)=60135\frac{ar^2(1+r)}{a(1+r)} = \frac{60}{135} r2=60135r^2 = \frac{60}{135}

    Simplify the fraction:

    r2=12×527×5=1227=4×39×3=49r^2 = \frac{12 \times 5}{27 \times 5} = \frac{12}{27} = \frac{4 \times 3}{9 \times 3} = \frac{4}{9}

    Take the square root of both sides:

    r=±49=±23r = \pm\sqrt{\frac{4}{9}} = \pm\frac{2}{3}

    Since the common ratio is positive, we take the positive value.

    r=23r = \frac{2}{3}

  2. Substitute the value of rr into Equation 1 to find the first term aa.

    a(1+r)=135a(1+r) = 135 a(1+23)=135a\left(1+\frac{2}{3}\right) = 135 a(33+23)=135a\left(\frac{3}{3}+\frac{2}{3}\right) = 135 a(53)=135a\left(\frac{5}{3}\right) = 135 a=135×35a = 135 \times \frac{3}{5} a=27×3a = 27 \times 3 a=81a = 81

    The common ratio is 23\boxed{\frac{2}{3}} and the first term is 81\boxed{81}.

Part ii243

Calculate the limit of the sum of the first nn terms as nn becomes large.

This refers to the sum to infinity of a geometric series, denoted as S∞S_\infty. The formula for the sum to infinity is S∞=a1−rS_\infty = \frac{a}{1-r}, provided that ∣r∣<1|r| < 1.

  1. Check the condition for the sum to infinity.

    We have r=23r = \frac{2}{3}.

    ∣r∣=∣23∣=23|r| = \left|\frac{2}{3}\right| = \frac{2}{3}

    Since 23<1\frac{2}{3} < 1, the sum to infinity exists.

  2. Calculate the sum to infinity using a=81a=81 and r=23r=\frac{2}{3}.

    S∞=a1−rS_\infty = \frac{a}{1-r} S∞=811−23S_\infty = \frac{81}{1-\frac{2}{3}} S∞=8133−23S_\infty = \frac{81}{\frac{3}{3}-\frac{2}{3}} S∞=8113S_\infty = \frac{81}{\frac{1}{3}} S∞=81×3S_\infty = 81 \times 3 S∞=243S_\infty = 243

    The limit of the sum of the first nn terms as nn becomes large is 243\boxed{243}.

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