Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Answer: d = 7, a = −32, so the 31st term = −32 + 30 × 7 = 178.
Step-by-step solution
Given: t11 = 38; t16 = 73
To find: t31
To find: t31
Idea: Write both given terms as a + (n − 1)d, subtract to get d, then find a and the 31st term.
- t11 = a + 10d = 38 … (1); t16 = a + 15d = 73 … (2).1 mark
- (2) − (1): 5d = 35 ⇒ d = 7. From (1): a = 38 − 70 = −32.1 mark
- t31 = a + 30d = −32 + 210 = 178.1 mark
The 31st term is 178.
Check: Shortcut: from the 16th to the 31st term is 15 steps: 73 + 15 × 7 = 73 + 105 = 178 ✓.
Answer to write in the exam
t11 = a + 10d = 38 … (1)
t16 = a + 15d = 73 … (2)
(2) − (1): 5d = 35 ⇒ d = 7
a = 38 − 70 = −32
t31 = a + 30d = −32 + 210 = 178
∴ 31st term = 178
Common mistakes that cost marks
- Writing the 11th term as a + 11d. It has 10 differences: a + 10d.
- Sign slip for a: 38 − 70 = −32, not 32.
- Dividing 73 − 38 = 35 by 16 − 11 = 5 correctly but then adding 31 × 7 to a (should be 30 × 7).
How this can come in the exam
MCQ (1 mark)
The 3rd term of an AP is 7 and the 8th term is 22. The common difference is
- 3
- 15
- 5
- 7
Show answer
(A) 3
5d = 22 − 7 = 15 ⇒ d = 3.
Short answer (3 marks)
The 7th term of an AP is 32 and its 13th term is 62. Find its 25th term.
Show answer
6d = 30 ⇒ d = 5 (1 mark); a = 32 − 30 = 2 (1 mark); 25th term = 2 + 24 × 5 = 122 (1 mark).Try one yourself
The 4th term of an AP is 10 and its 10th term is −8. Find its 20th term.
Show answer
6d = −18 ⇒ d = −3; a = 19; 20th term = 19 − 57 = −38.
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