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Determine which of the following sequences are arithmetic progressions and find the next three terms for each AP: (i) 2, 4, 8, 16,... (ii) 2, 5/2, 3, 7/2,... (iii) -1.2, -3.2, -5.2, -7.2,...

Mathematics

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Let's determine which of these sequences are Arithmetic Progressions (APs). An AP has a constant difference between consecutive terms.

Part i· 1 step

2,4,8,16,…2, 4, 8, 16, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=4−2=2a_2 - a_1 = 4 - 2 = 2
    a3−a2=8−4=4a_3 - a_2 = 8 - 4 = 4
    Since the differences are not constant (2≠42 \neq 4), this sequence is not an AP.

Part ii· 3 steps

2,52,3,72,…2, \frac{5}{2}, 3, \frac{7}{2}, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=52−2=52−42=12a_2 - a_1 = \frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2}
    a3−a2=3−52=62−52=12a_3 - a_2 = 3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2}
    a4−a3=72−3=72−62=12a_4 - a_3 = \frac{7}{2} - 3 = \frac{7}{2} - \frac{6}{2} = \frac{1}{2}

  2. Determine if it's an AP and find the common difference.

    The common difference is constant, d=12d = \frac{1}{2}. So, this is an AP.

  3. Write the next three terms.

    a5=72+12=82=4a_5 = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4
    a6=4+12=92a_6 = 4 + \frac{1}{2} = \frac{9}{2}
    a7=92+12=102=5a_7 = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5
    This is an AP with d=12d = \boxed{\frac{1}{2}}. The next three terms are 4,92,5\boxed{4, \frac{9}{2}, 5}.

Part iii· 3 steps

−1.2,−3.2,−5.2,−7.2,…-1.2, -3.2, -5.2, -7.2, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=−3.2−(−1.2)=−3.2+1.2=−2.0a_2 - a_1 = -3.2 - (-1.2) = -3.2 + 1.2 = -2.0
    a3−a2=−5.2−(−3.2)=−5.2+3.2=−2.0a_3 - a_2 = -5.2 - (-3.2) = -5.2 + 3.2 = -2.0
    a4−a3=−7.2−(−5.2)=−7.2+5.2=−2.0a_4 - a_3 = -7.2 - (-5.2) = -7.2 + 5.2 = -2.0

  2. Determine if it's an AP and find the common difference.

    The common difference is constant, d=−2.0d = -2.0. So, this is an AP.

  3. Write the next three terms.

    a5=−7.2+(−2.0)=−9.2a_5 = -7.2 + (-2.0) = -9.2
    a6=−9.2+(−2.0)=−11.2a_6 = -9.2 + (-2.0) = -11.2
    a7=−11.2+(−2.0)=−13.2a_7 = -11.2 + (-2.0) = -13.2
    This is an AP with d=−2.0d = \boxed{-2.0}. The next three terms are −9.2,−11.2,−13.2\boxed{-9.2, -11.2, -13.2}.

Part iv· 3 steps

−10,−6,−2,2,…-10, -6, -2, 2, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=−6−(−10)=−6+10=4a_2 - a_1 = -6 - (-10) = -6 + 10 = 4
    a3−a2=−2−(−6)=−2+6=4a_3 - a_2 = -2 - (-6) = -2 + 6 = 4
    a4−a3=2−(−2)=2+2=4a_4 - a_3 = 2 - (-2) = 2 + 2 = 4

  2. Determine if it's an AP and find the common difference.

    The common difference is constant, d=4d = 4. So, this is an AP.

  3. Write the next three terms.

    a5=2+4=6a_5 = 2 + 4 = 6
    a6=6+4=10a_6 = 6 + 4 = 10
    a7=10+4=14a_7 = 10 + 4 = 14
    This is an AP with d=4d = \boxed{4}. The next three terms are 6,10,14\boxed{6, 10, 14}.

Part v· 3 steps

3,3+2,3+22,3+32,…3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=(3+2)−3=2a_2 - a_1 = (3+\sqrt{2}) - 3 = \sqrt{2}
    a3−a2=(3+22)−(3+2)=3+22−3−2=2a_3 - a_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = 3+2\sqrt{2} - 3 - \sqrt{2} = \sqrt{2}
    a4−a3=(3+32)−(3+22)=3+32−3−22=2a_4 - a_3 = (3+3\sqrt{2}) - (3+2\sqrt{2}) = 3+3\sqrt{2} - 3 - 2\sqrt{2} = \sqrt{2}

  2. Determine if it's an AP and find the common difference.

    The common difference is constant, d=2d = \sqrt{2}. So, this is an AP.

  3. Write the next three terms.

    a5=(3+32)+2=3+42a_5 = (3+3\sqrt{2}) + \sqrt{2} = 3+4\sqrt{2}
    a6=(3+42)+2=3+52a_6 = (3+4\sqrt{2}) + \sqrt{2} = 3+5\sqrt{2}
    a7=(3+52)+2=3+62a_7 = (3+5\sqrt{2}) + \sqrt{2} = 3+6\sqrt{2}
    This is an AP with d=2d = \boxed{\sqrt{2}}. The next three terms are 3+42,3+52,3+62\boxed{3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}}.

Part vi· 1 step

0.2,0.22,0.222,0.2222,…0.2, 0.22, 0.222, 0.2222, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=0.22−0.2=0.02a_2 - a_1 = 0.22 - 0.2 = 0.02
    a3−a2=0.222−0.22=0.002a_3 - a_2 = 0.222 - 0.22 = 0.002
    Since the differences are not constant (0.02≠0.0020.02 \neq 0.002), this sequence is not an AP.

Part vii· 3 steps

0,−4,−8,−12,…0, -4, -8, -12, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=−4−0=−4a_2 - a_1 = -4 - 0 = -4
    a3−a2=−8−(−4)=−8+4=−4a_3 - a_2 = -8 - (-4) = -8 + 4 = -4
    a4−a3=−12−(−8)=−12+8=−4a_4 - a_3 = -12 - (-8) = -12 + 8 = -4

  2. Determine if it's an AP and find the common difference.

    The common difference is constant, d=−4d = -4. So, this is an AP.

  3. Write the next three terms.

    a5=−12+(−4)=−16a_5 = -12 + (-4) = -16
    a6=−16+(−4)=−20a_6 = -16 + (-4) = -20
    a7=−20+(−4)=−24a_7 = -20 + (-4) = -24
    This is an AP with d=−4d = \boxed{-4}. The next three terms are −16,−20,−24\boxed{-16, -20, -24}.

Part viii· 3 steps

−12,−12,−12,−12,…-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots

  1. Calculate the differences between consecutive terms.

    a2−a1=−12−(−12)=0a_2 - a_1 = -\frac{1}{2} - (-\frac{1}{2}) = 0
    a3−a2=−12−(−12)=0a_3 - a_2 = -\frac{1}{2} - (-\frac{1}{2}) = 0
    a4−a3=−12−(−12)=0a_4 - a_3 = -\frac{1}{2} - (-\frac{1}{2}) = 0

  2. Determine if it's an AP and find the common difference.

    The common difference is constant, d=0d = 0. So, this is an AP.

  3. Write the next three terms.

    a5=−12+0=−12a_5 = -\frac{1}{2} + 0 = -\frac{1}{2}
    a6=−12+0=−12a_6 = -\frac{1}{2} + 0 = -\frac{1}{2}
    a7=−12+0=−12a_7 = -\frac{1}{2} + 0 = -\frac{1}{2}
    This is an AP with d=0d = \boxed{0}. The next three terms are −12,−12,−12\boxed{-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}}.

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