Math Problem Statement

a) The 8th term of an arithmetic series is 57 and the 17th term is 111. Required: i. Determine the common difference ii. The first term iii. The sum of the first 28 terms b) The 10th term of a geometric progression is 177,147 and the 6th term of the series is 2,187. Required: i. The common ratio ii. The value of the first term

Solution

Let's solve the given problems step by step.


(a) Arithmetic Progression

The general term for an arithmetic sequence is:

an=a+(n1)da_n = a + (n-1)d

where:

  • aa = first term
  • dd = common difference
  • ana_n = nn-th term

Given:

  • a8=57a_8 = 57
  • a17=111a_{17} = 111

i. Find the Common Difference (dd)

From the formula: a8=a+7danda17=a+16da_8 = a + 7d \quad \text{and} \quad a_{17} = a + 16d

Subtract the two equations: (a+16d)(a+7d)=11157(a + 16d) - (a + 7d) = 111 - 57 9d=54d=69d = 54 \quad \Rightarrow \quad d = 6


ii. Find the First Term (aa)

Substitute d=6d = 6 into a8=a+7da_8 = a + 7d: 57=a+7(6)57 = a + 7(6) 57=a+42a=1557 = a + 42 \quad \Rightarrow \quad a = 15


iii. Sum of the First 28 Terms

The sum of the first nn terms of an arithmetic progression is:

Sn=n2(2a+(n1)d)S_n = \frac{n}{2} \big(2a + (n-1)d\big)

For n=28n = 28: S28=282(2(15)+(281)(6))S_{28} = \frac{28}{2} \big(2(15) + (28-1)(6)\big) S28=14(30+27(6))S_{28} = 14 \big(30 + 27(6)\big) S28=14(30+162)=14(192)=2688S_{28} = 14 \big(30 + 162\big) = 14(192) = 2688


(b) Geometric Progression

The general term for a geometric sequence is:

an=arn1a_n = ar^{n-1}

where:

  • aa = first term
  • rr = common ratio
  • ana_n = nn-th term

Given:

  • a10=177,147a_{10} = 177,147
  • a6=2,187a_6 = 2,187

i. Find the Common Ratio (rr)

From the formula: a10=ar9anda6=ar5a_{10} = ar^9 \quad \text{and} \quad a_6 = ar^5

Divide the two equations: a10a6=ar9ar5r4=177,1472,187\frac{a_{10}}{a_6} = \frac{ar^9}{ar^5} \quad \Rightarrow \quad r^4 = \frac{177,147}{2,187}

Simplify: r4=81r=814=3r^4 = 81 \quad \Rightarrow \quad r = \sqrt[4]{81} = 3


ii. Find the First Term (aa)

Substitute r=3r = 3 into a6=ar5a_6 = ar^5: 2,187=a(35)2,187 = a(3^5) 2,187=a(243)a=2,187243=92,187 = a(243) \quad \Rightarrow \quad a = \frac{2,187}{243} = 9


Final Answers

(a) Arithmetic Progression

  1. d=6d = 6
  2. a=15a = 15
  3. S28=2688S_{28} = 2688

(b) Geometric Progression

  1. r=3r = 3
  2. a=9a = 9

Would you like a detailed explanation of any step or related examples?


Expanding Questions:

  1. How would you calculate the nn-th term of an arithmetic sequence with different values of nn?
  2. Can you derive the formula for the sum of a geometric series?
  3. How would the solution change if the given terms in the sequences were decimals?
  4. What is the relationship between rr and the growth rate in a geometric sequence?
  5. How does changing dd or rr affect the behavior of the sequences?

Tip: When working with series, always ensure the values substituted for nn, aa, dd, and rr match their respective equations to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression
Geometric Progression

Formulas

Arithmetic sequence formula: a_n = a + (n-1)d
Sum of arithmetic series: S_n = (n/2) * [2a + (n-1)d]
Geometric sequence formula: a_n = ar^(n-1)

Theorems

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Suitable Grade Level

Grades 9-12