IB Mathematics SL 1.2 Arithmetic sequences and series AI HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
a) To find the common difference of the sequence:
\( d = -7 \)
Thus:
The common difference is \( -7 \) [1]
b) To calculate the sum of the first \( 50 \) terms:
Use the sum formula:
\( S_{50} = \frac{50}{2} \times (2 \times 124 + 49 \times (-7)) \)
Calculate:
\( S_{50} = -2375 \)
Thus:
The sum is \( -2375 \) [2]
c) To find the value of \( k \) where \( u_k \) is negative:
Solve the inequality:
\( 124 – 7 \times (k – 1) < 0 \)
Simplify:
\( k > 18.7 \)
Since \( k \) is an integer:
\( k = 19 \)
Thus:
The value of \( k \) is \( 19 \) [3]
▶️ Answer/Explanation
a)(i) To find the common difference:
Set up the equations:
\( u_5 = u_1 + 4d = 20 \)
\( u_{12} = u_1 + 11d = 41 \)
Subtract to solve:
\( 7d = 21 \)
\( d = 3 \)
Thus:
The common difference is \( 3 \) [2]
(ii) To find the first term:
Using \( u_5 \):
\( u_1 + 12 = 20 \)
\( u_1 = 8 \)
Thus:
The first term is \( 8 \) [1]
b) To calculate the eighty-fourth term:
Use the formula:
\( u_{84} = 8 + (84 – 1) \times 3 \)
\( = 257 \)
Thus:
The eighty-fourth term is \( 257 \) [1]
c) To calculate the sum of the first \( 200 \) terms:
Use the sum formula:
\( S_{200} = 100 \times (16 + 199 \times 3) \)
\( = 61300 \)
Thus:
The sum is \( 61300 \) [2]