Algebra — Sequences · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.
1. The sum of the first n terms of an AP with first term a and common difference d is given by which formula?
- AS = 2n[2a + (n-1)d]
- BS = 2n[a + (n-1)d]
- CS = n[2a + (n-1)d]
- DS = 2n[2a + nd]
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Answer: A) S = 2n[2a + (n-1)d]
The standard formula for the sum of n terms of an AP is S_n = 2n[2a + (n-1)d].
3. The sum of the first n terms of an AP is given by S_n = 4n - n².
Find: (a) the first term, and (b) the sum of the first 2 terms.
Which option gives both correctly?
- AFirst term = 3, S_2 = 4
- BFirst term = 4, S_2 = 4
- CFirst term = 3, S_2 = 5
- DFirst term = 4, S_2 = 0
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Answer: A) First term = 3, S_2 = 4
S_1 = 4(1) - 1 = 3 = first term. S_2 = 4(2) - 4 = 8 - 4 = 4.
4. Find the sum of all odd numbers lying between 20 and 80.
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Answer: B) 1500
The odd numbers are 21, 23, ..., 79, so n = 279−21 + 1 = 30. S = 230(21 + 79) = 15(100) = 1500. The value 1450 uses n = 29 and 1550 uses n = 31 — both miscount the terms. 3000 drops the factor 21. The shortcut "sum of the first n odd numbers = n²" does not apply here, because this list does not start at 1.
5. An AP has 15 terms whose total is 600, and its first term is 12.
Find the sum of its last 5 terms.
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Answer: C) 300
First find d: 600 = 215[24 + 14d], so 80 = 24 + 14d, giving 14d = 56 and d = 4. The last 5 terms are terms 11 to 15, so their sum is S_15 - S_10 = 600 - 210[24 + 36] = 600 - 300 = 300. (Check: a_11 = 52, a_15 = 68 and 25(52 + 68) = 300.) The value 248 subtracts S_11, wrongly deleting the 11th term too. 290 averages a_10 and a_15 over 5 terms, but a_10 is not one of the last five. 200 assumes the last third of the terms carries a third of the total, which is false whenever d is not 0.