These eight original questions follow the order of the lessons in progressions. Begin each solution by stating the type of progression and the values of a, d or r, and n.
Questions
Question 1
An arithmetic progression is 3, 8, 13, … Find T₁₅ and S₁₅.
Answer
a = 3, d = 5, n = 15.
T₁₅ = 3 + 14 × 5 = 3 + 70 = 73.
S₁₅ = 15/2 × (3 + 73) = 15/2 × 76 = 15 × 38 = 570.
Question 2
In an arithmetic progression, T₃ = 11 and T₈ = 31. Find a and d, then find S₁₀.
Answer
T₈ − T₃ = 5d, so 31 − 11 = 5d and d = 4.
T₃ = a + 2d = 11, so a + 8 = 11 and a = 3.
S₁₀ = 10/2 × [2(3) + 9(4)] = 5 × (6 + 36) = 5 × 42 = 210.
Check: T₈ = 3 + 7 × 4 = 31.
Question 3
A geometric progression is 2, 6, 18, … Find T₆ and S₆.
Answer
a = 2, r = 3, n = 6.
T₆ = 2 × 3⁵ = 2 × 243 = 486.
S₆ = 2(3⁶ − 1) ÷ (3 − 1) = 2 × 728 ÷ 2 = 728.
Check by adding: 2 + 6 + 18 + 54 + 162 + 486 = 728.
Question 4
In a geometric progression, T₃ = 20 and T₆ = 160. Find r and a, then find S₅.
Answer
T₆ ÷ T₃ = r³, so r³ = 160 ÷ 20 = 8 and r = 2.
T₃ = ar² = 20, so 4a = 20 and a = 5.
S₅ = 5(2⁵ − 1) ÷ (2 − 1) = 5 × 31 = 155.
Check: T₆ = 5 × 2⁵ = 160.
Question 5
Find the sum to infinity of the geometric series 18 + 6 + 2 + …
Answer
r = 6 ÷ 18 = 1/3, and |1/3| < 1, so the sum exists.
S∞ = 18 ÷ (1 − 1/3) = 18 ÷ (2/3) = 18 × 3/2 = 27.
Question 6
A geometric series has first term 12 and second term 4k. (a) Find the range of k for which the sum to infinity exists. (b) Find the sum to infinity when k = 1.5.
Answer
(a) r = 4k ÷ 12 = k/3. The condition is −1 < k/3 < 1, so −3 < k < 3.
(b) When k = 1.5, r = 0.5. S∞ = 12 ÷ (1 − 0.5) = 12 ÷ 0.5 = 24.
Question 7
A student deposits RM150 in January. Each month she deposits RM10 more than the month before. Find her total deposits by December.
Answer
The same amount is added each month, so this is arithmetic with a = 150, d = 10, n = 12.
S₁₂ = 12/2 × [2(150) + 11(10)] = 6 × (300 + 110) = 6 × 410 = RM2 460.
Check: December is T₁₂ = 150 + 110 = 260, and 6 × (150 + 260) = 2 460.
Question 8
A fictional pond plant covers 8 m² now and grows by 25% each week. (a) Find the area after 4 weeks. (b) Find the smallest number of whole weeks after which the area first exceeds 30 m².
Answer
A 25% rise multiplies by 1.25, so this is geometric with a = 8 and r = 1.25. The area now is T₁, so “after n weeks” is T_(n+1).
(a) After 4 weeks: T₅ = 8 × 1.25⁴ = 8 × 2.441 4 = 19.53, so 19.53 m² (to 2 decimal places).
(b) Solve 8 × 1.25ⁿ > 30, so 1.25ⁿ > 3.75.
Taking logarithms: n > log 3.75 ÷ log 1.25 = 5.92.
Check: after 5 weeks the area is 8 × 1.25⁵ = 24.41, which is below 30. After 6 weeks it is 8 × 1.25⁶ = 30.52, which is above. So 6 weeks.
If you got these wrong
| Question | Skill to revisit |
|---|---|
| 1 and 2 | Terms and sums of arithmetic progressions |
| 3 and 4 | Terms and sums of geometric progressions |
| 5 and 6 | The sum to infinity condition |
| 7 and 8 | Turning payment and growth stories into progressions |
Keep a record of each slip in the mistake log and paper-error review. If the same type of slip keeps returning, see online one-to-one Additional Mathematics tuition.