Question 1 The first and last terms in each arithmetic series are given. Determine the sum of the series. (a) a = 21 and t9 = 4, (b) a = 11 and t20 = 101. Solution The sum of an arithmetic series is given by: Sn = n (a + tn ), 2 where: • n is the number of terms, • a is the first term, • tn is the last term. (a) Solution: Given a = 12 , t9 = 4, and n = 9: 9 S9 = 2 1 +4 . 2 Simplify inside the parentheses: 1 1 8 9 +4= + = . 2 2 2 2 Substitute: S9 = 81 9 9 · = . 2 2 4 S9 = 81 or 20.25. 4 Final Answer: (b) Solution: Given a = 11, t20 = 101, and n = 20: S20 = 20 (11 + 101). 2 1 Simplify: S20 = 10 · 112 = 1120. Final Answer: S20 = 1120. Question 2 Determine the sum of the arithmetic series: 1 + 2 + 5 + · · · + 164. Solution The general formula for the n-th term of an arithmetic sequence is: tn = a + (n − 1)d, where a is the first term, d is the common difference, and n is the number of terms. Step 1: Find the number of terms (n) Here, a = 1, d = 3, and tn = 164: 164 = 1 + (n − 1) · 3. Simplify: 164 = 1 + 3n − 3, 164 = 3n − 2, 166 = 3n, 166 = 55. n= 3 Step 2: Use the sum formula The sum of an arithmetic series is: n Sn = (a + tn ). 2 Substitute n = 55, a = 1, and tn = 164: S55 = 55 (1 + 164), 2 2 S55 = 55 · 165 = 55 · 82.5 = 4537.5. 2 Final Answer: S55 = 4537.5. Question 3 The 15th term in an arithmetic sequence is 43, and the sum of the first 15 terms of the series is 120. Determine the first three terms of the sequence. Solution The formula for the n-th term of an arithmetic sequence is: tn = a + (n − 1)d. The formula for the sum of the first n terms is: Sn = n (2a + (n − 1)d) . 2 Step 1: Set up the equations From t15 = 43: 43 = a + 14d. From S15 = 120: (1) 15 (2a + 14d) , 2 240 = 15(2a + 14d), 120 = 16 = 2a + 14d. Step 2: Solve the system of equations From equation (1): a = 43 − 14d. Substitute a = 43 − 14d into equation (2): 16 = 2(43 − 14d) + 14d, 16 = 86 − 28d + 14d, 3 (2) 16 = 86 − 14d, 14d = 86 − 16 = 70, d = 5. Substitute d = 5 into a = 43 − 14d: a = 43 − 14(5) = 43 − 70 = −27. Step 3: Find the first three terms First term: a = −27, Second term: a + d = −27 + 5 = −22, Third term: a + 2d = −27 + 2(5) = −17. Final Answer: The first three terms are − 27, −22, −17. Question 6 Determine the sum of the geometric series: 4 128 1 2 + + + ··· + . 3 9 27 6561 Solution The formula for the sum of a finite geometric series is: Sn = a 1 − rn , 1−r where: • a= 1 3 is the first term, • r= 2 3 is the common ratio, • n = 8 is the number of terms. 4 Substitute into the formula: S8 = 1 3 · (1 − 1− 2 8 ) 3 2 3 . Simplify: 256 1 1 − 6561 · , 1 3 3 256 S8 = 1 − , 6561 256 6305 6561 − = . S8 = 6561 6561 6561 S8 = Final Answer: S8 = 6305 . 6561 Question 13 Which term of the geometric sequence 1, 3, 9, . . . has a value of 19, 683? Solution The formula for the n-th term of a geometric sequence is: tn = a · rn−1 . Given a = 1, r = 3, and tn = 19, 683: 19, 683 = 1 · 3n−1 . Simplify: 19, 683 = 3n−1 . Recognize that 19, 683 = 39 : 3n−1 = 39 . Equating exponents: n − 1 = 9, n = 10. Final Answer: The term with a value of 19, 683 is the 10th term. 5