📝 Chapter 5: Sequences and Series

Assessment AS Learning — Practice Quiz
🔄 Not Graded — Unlimited Retakes
Purpose: Self-check arithmetic and geometric sequences and series.
Score: 0 / 12
Arithmetic Sequences & Series
Question 1
For the arithmetic sequence \( 5, 8, 11, 14, \ldots \), find \( t_{20} \).
Solution:
\( a = 5, d = 3 \). \( t_{20} = 5 + (20 - 1)(3) = 5 + 57 = 62 \).
Question 2
Which term of \( 7, 11, 15, 19, \ldots \) equals 95?
Solution:
\( 95 = 7 + (n - 1)(4) \Rightarrow n = 23 \).
Question 3
Find \( S_{15} \) for \( 4 + 7 + 10 + 13 + \ldots \).
Solution:
\( a = 4, d = 3 \). \( S_{15} = \tfrac{15}{2}(2(4) + 14(3)) = \tfrac{15}{2}(50) = 375 \).
Question 4
Sum the first 50 even numbers (2 + 4 + ... + 100).
Solution:
\( a = 2, d = 2, n = 50 \), \( t_{50} = 100 \). \( S_{50} = \tfrac{50}{2}(2 + 100) = 25(102) = 2550 \).
Geometric Sequences & Series
Question 5
For the geometric sequence \( 2, 6, 18, 54, \ldots \), find \( t_8 \).
Solution:
\( a = 2, r = 3 \). \( t_8 = 2 \cdot 3^7 = 2 \cdot 2187 = 4374 \).
Question 6
A geometric sequence has \( t_1 = 5 \) and \( t_4 = 40 \). Find \( r \).
Solution:
\( t_4 = 5 r^3 = 40 \Rightarrow r^3 = 8 \Rightarrow r = 2 \).
Question 7
Find \( S_6 \) for \( 3, 6, 12, 24, \ldots \).
Solution:
\( a = 3, r = 2 \). \( S_6 = \tfrac{3(2^6 - 1)}{2 - 1} = 3 \cdot 63 = 189 \).
Question 8
Sum: \( 1 + 2 + 4 + 8 + \ldots + 1024 \). (Hint: 1024 = \( 2^{10} \), so 11 terms.)
Solution:
\( a = 1, r = 2, n = 11 \). \( S_{11} = \tfrac{1(2^{11} - 1)}{1} = 2047 \).
Recursive & Mixed
Question 9
For \( t_n = t_{n-1} + 4 \) with \( t_1 = 7 \), find \( t_5 \).
Solution:
Arithmetic with \( a = 7, d = 4 \). \( t_5 = 7 + 4(4) = 23 \).
Question 10
Sigma: \( \displaystyle\sum_{k=1}^{4} (2k + 3) \) equals:
Solution:
5 + 7 + 9 + 11 = 32.
Question 11
Identify whether \( 100, 80, 64, 51.2, \ldots \) is arithmetic, geometric, or neither.
Solution:
\( 80/100 = 0.8, 64/80 = 0.8, 51.2/64 = 0.8 \). Constant ratio \( r = 0.8 \). Geometric.
Question 12
A ball dropped from 10 m bounces 80% as high each time. Total distance travelled before the 5th bounce (down + up between bounces, in m):
Solution:
Down: 10 + 8 + 6.4 + 5.12 + 4.096. Up after each bounce: 8 + 6.4 + 5.12 + 4.096. Total \( \approx 33.6 + 23.6 = 39.5 \) m (approximately).