Detailed Formulas, Definitions, and Step-by-Step Solutions
A sequence is an ordered list of numbers. An arithmetic sequence has a common difference, while a geometric sequence has a common ratio. A series is the sum of the terms of a sequence.
Common Difference (d): \( d = a_{n} - a_{n-1} \)
Nth Term:
\[ a_n = a + (n-1)d \]Sum of \( n \) terms (\( S_n \)):
\[ S_n = \frac{n}{2}[2a + (n-1)d] \] \[ S_n = \frac{n}{2}(a + a_n) \]Common Ratio (r): \( r = \frac{a_{n}}{a_{n-1}} \)
Nth Term:
\[ a_n = a \cdot r^{n-1} \]Sum of \( n \) terms (\( S_n \)):
\[ S_n = \frac{a(r^n - 1)}{r - 1}, \quad r \ne 1 \]An infinite geometric series has a finite sum (converges) only if the common ratio satisfies \( |r| < 1 \). If \( |r| \ge 1 \), the series diverges and has no sum.
Sum to Infinity (\( S_{\infty} \)):
\[ S_{\infty} = \frac{a}{1 - r} \]Question 1 (Arithmetic): Find the 15th term of the arithmetic sequence: \( 10, 6, 2, -2, \dots \)
Identify the first term \( a = 10 \) and the common difference \( d = 6 - 10 = -4 \).
\[ a_{15} = 10 + (15-1)(-4) \] \[ a_{15} = 10 + 14(-4) = 10 - 56 = -46 \]Question 2 (Geometric): Find the 8th term of the geometric sequence: \( 5, 10, 20, 40, \dots \)
Identify \( a = 5 \) and \( r = 10/5 = 2 \). Solve for \( a_8 \):
\[ a_8 = 5 \cdot (2)^{8-1} = 5 \cdot 2^7 \] \[ a_8 = 5 \cdot 128 = 640 \]Question 3 (Challenge): In an arithmetic sequence, \( a_3 = 13 \) and \( a_9 = 37 \). Find the first term \( a \) and the common difference \( d \).
Set up two equations using \( a_n = a + (n-1)d \):
(1) \( a + 2d = 13 \)
(2) \( a + 8d = 37 \)
Subtract (1) from (2): \( 6d = 24 \Rightarrow d = 4 \).
Substitute \( d=4 \) into (1): \( a + 2(4) = 13 \Rightarrow a + 8 = 13 \Rightarrow a = 5 \).
Question 4 (Arithmetic Sum): Find the sum of the first 50 even positive integers.
The sequence is \( 2, 4, 6, \dots \), so \( a = 2 \), \( d = 2 \), and \( n = 50 \).
\[ S_{50} = \frac{50}{2}[2(2) + (50-1)(2)] \] \[ S_{50} = 25[4 + 98] = 25[102] = 2550 \]Question 5 (Geometric Sum): Find the sum of the first 6 terms of the geometric series: \( 3, -6, 12, -24, \dots \)
Identify \( a = 3 \), \( r = -2 \), and \( n = 6 \):
\[ S_6 = \frac{3((-2)^6 - 1)}{-2 - 1} \] \[ S_6 = \frac{3(64 - 1)}{-3} = \frac{3(63)}{-3} = -63 \]Question 6 (Challenge): Find the sum of all three-digit multiples of 7.
First multiple \( a = 105 \). Last multiple \( a_n = 994 \). Common difference \( d = 7 \).
Find \( n \): \( 994 = 105 + (n-1)7 \Rightarrow 889 = 7(n-1) \Rightarrow n = 128 \).
Find Sum: \[ S_{128} = \frac{128}{2}(105 + 994) = 64(1099) = 70336 \]
Question 7: Determine the sum of the infinite series: \( 1 + \frac{1}{2} + \frac{1}{4} + \dots \)
Identify \( a = 1 \) and \( r = 0.5 \). Since \( |0.5| < 1 \), the series converges:
\[ S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - 0.5} = \frac{1}{0.5} = 2 \]Question 8: An infinite geometric series has a sum of 20 and a first term of 5. Find the common ratio \( r \).