Math Problem Statement

Find the nth term and the sum of the following arithmetic series and geometric series.

Solution

Let's work through the questions from the uploaded image.

Arithmetic Series

The general formula for the nth term ana_n and sum SnS_n of an arithmetic series are:

  • an=a1+(n1)da_n = a_1 + (n-1)d
  • Sn=n2(a1+an)S_n = \frac{n}{2} \cdot (a_1 + a_n)

Where:

  • a1a_1 is the first term
  • dd is the common difference
  • nn is the number of terms

1. 7,11,15,7, 11, 15, \dots

  • Common difference d=117=4d = 11 - 7 = 4

  • Find a7a_7:
    a7=7+(71)×4=7+24=31a_7 = 7 + (7-1) \times 4 = 7 + 24 = 31

  • Find S7S_7:
    S7=72×(7+31)=72×38=133S_7 = \frac{7}{2} \times (7 + 31) = \frac{7}{2} \times 38 = 133

2. 18,11,4,18, 11, 4, \dots

  • Common difference d=1118=7d = 11 - 18 = -7

  • Find a6a_6:
    a6=18+(61)×(7)=1835=17a_6 = 18 + (6-1) \times (-7) = 18 - 35 = -17

  • Find S6S_6:
    S6=62×(18+(17))=3×1=3S_6 = \frac{6}{2} \times (18 + (-17)) = 3 \times 1 = 3

3. 7,5,3,-7, -5, -3, \dots

  • Common difference d=5(7)=2d = -5 - (-7) = 2

  • Find a23a_{23}:
    a23=7+(231)×2=7+44=37a_{23} = -7 + (23-1) \times 2 = -7 + 44 = 37

  • Find S23S_{23}:
    S23=232×(7+37)=232×30=345S_{23} = \frac{23}{2} \times (-7 + 37) = \frac{23}{2} \times 30 = 345

4. 3,312,4,3, 3 \frac{1}{2}, 4, \dots

  • Common difference d=3.53=0.5d = 3.5 - 3 = 0.5

  • Find a16a_{16}:
    a16=3+(161)×0.5=3+7.5=10.5a_{16} = 3 + (16-1) \times 0.5 = 3 + 7.5 = 10.5

  • Find S16S_{16}:
    S16=162×(3+10.5)=8×13.5=108S_{16} = \frac{16}{2} \times (3 + 10.5) = 8 \times 13.5 = 108

5. 17,12,7,-17, -12, -7, \dots

  • Common difference d=12(17)=5d = -12 - (-17) = 5

  • Find a24a_{24}:
    a24=17+(241)×5=17+115=98a_{24} = -17 + (24-1) \times 5 = -17 + 115 = 98

  • Find S24S_{24}:
    S24=242×(17+98)=12×81=972S_{24} = \frac{24}{2} \times (-17 + 98) = 12 \times 81 = 972


Geometric Series

The general formula for the nth term ana_n and sum SnS_n of a geometric series are:

  • an=a1×rn1a_n = a_1 \times r^{n-1}
  • Sn=a1×1rn1rS_n = a_1 \times \frac{1 - r^n}{1 - r} (for r<1|r| < 1)
  • Sum of infinite series S=a11rS_{\infty} = \frac{a_1}{1 - r} (for r<1|r| < 1)

1. 1,6,36,216,-1, 6, -36, 216, \dots

  • Common ratio r=61=6r = \frac{6}{-1} = -6

  • Find a19a_{19}:
    a19=1×(6)18=618a_{19} = -1 \times (-6)^{18} = -6^{18} a19a_{19} is a very large value.

  • Find S19S_{19} using the sum formula for a geometric series (not suitable for this large common ratio).

2. 12,18,132,\frac{1}{2}, \frac{1}{8}, \frac{1}{32}, \dots

  • Common ratio r=1/81/2=14r = \frac{1/8}{1/2} = \frac{1}{4}

  • Find a27a_{27}:
    a27=12×(14)26a_{27} = \frac{1}{2} \times \left(\frac{1}{4}\right)^{26}

  • Find S27S_{27} and infinite sum (since r<1|r| < 1) using geometric formulas.


Would you like detailed calculations for the geometric series or any other part?

Here are 5 related questions:

  1. What is the nth term formula for a quadratic sequence?
  2. Can the sum formula for an arithmetic series apply to negative sequences?
  3. What happens if the common difference in an arithmetic sequence is fractional?
  4. How do you determine the common ratio for fractional geometric sequences?
  5. What is the application of infinite series sums in real-life scenarios?

Tip: In geometric series, ensure the common ratio's magnitude is less than 1 to use the infinite sum formula effectively.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Series
Geometric Series

Formulas

nth term of arithmetic series: a_n = a_1 + (n-1) * d
Sum of arithmetic series: S_n = n/2 * (a_1 + a_n)
nth term of geometric series: a_n = a_1 * r^(n-1)
Sum of geometric series (finite): S_n = a_1 * (1 - r^n) / (1 - r) for |r| < 1
Sum of infinite geometric series: S_infinity = a_1 / (1 - r) for |r| < 1

Theorems

Arithmetic series sum formula
Geometric series sum formula

Suitable Grade Level

Grades 10-12