The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Answer: d = 5, a = −13: the first three terms are −13, −8, −3.
Step-by-step solution
Idea: Write each term as a + (n − 1)d, make two equations in a and d, and subtract.
- t4 + t8 = (a + 3d) + (a + 7d) = 2a + 10d = 24 … (1)½ mark
- t6 + t10 = (a + 5d) + (a + 9d) = 2a + 14d = 44 … (2)½ mark
- (2) − (1): 4d = 20 ⇒ d = 5. From (1): 2a = 24 − 50 = −26 ⇒ a = −13.1 mark
- First three terms: −13, −13 + 5 = −8, −8 + 5 = −3, i.e. −13, −8, −3.1 mark
The first three terms are −13, −8, −3.
Check: t₄ = 2, t₈ = 22: 2 + 22 = 24 ✓; t₆ = 12, t₁₀ = 32: 12 + 32 = 44 ✓.
Answer to write in the exam
t4 + t8 = 2a + 10d = 24 … (1)
t6 + t10 = 2a + 14d = 44 … (2)
(2) − (1): 4d = 20 ⇒ d = 5
2a = 24 − 50 = −26 ⇒ a = −13
∴ First three terms: −13, −8, −3
Common mistakes that cost marks
- Writing the 4th term as a + 4d. Use a + (n − 1)d: a + 3d.
- Sign error: 2a = 24 − 50 = −26, so a is negative.
- Giving a and d only. The question asks for the first three terms.
How this can come in the exam
MCQ (1 mark)
In an AP, t2 + t6 = 20 and t3 + t7 = 26. The common difference is
- 6
- 3
- 2
- 13
Show answer
(B) 3
Each term in the second sum is d more: 2d = 6 ⇒ d = 3.
Try one yourself
The sum of the 3rd and 7th terms of an AP is 6 and the sum of the 5th and 9th terms is 22. Find the first three terms.
Show answer
2a + 8d = 6, 2a + 12d = 22 ⇒ d = 4, a = −13. Terms: −13, −9, −5.
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