An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
(Hint: If ‘a’ is the first term and ‘d’ the common difference, then we arrive at the equations a + 2d = 12 and a + 49d = 106. Solve this pair of linear equations for ‘a’ and ‘d’.)
Answer: a = 8, d = 2, so the 29th term = 8 + 28 × 2 = 64.
Step-by-step solution
Given: 50 terms; t3 = 12; t50 = 106 (the last term)
To find: t29
To find: t29
Idea: The last of 50 terms is the 50th term. Write both given terms using a + (n − 1)d, subtract the equations to find d, then find a.
- t3 = a + 2d = 12 … (1); t50 = a + 49d = 106 … (2).1 mark
- (2) − (1): 47d = 94 ⇒ d = 2. Then a = 12 − 2 × 2 = 8.1 mark
- t29 = a + 28d = 8 + 28 × 2 = 64.1 mark
The 29th term is 64.
Check: t₃ = 8 + 4 = 12 ✓, t₅₀ = 8 + 98 = 106 ✓.
Answer to write in the exam
t3 = a + 2d = 12 … (1)
t50 = a + 49d = 106 … (2)
(2) − (1): 47d = 94 ⇒ d = 2
a = 12 − 4 = 8
t29 = a + 28d = 8 + 56 = 64
∴ 29th term = 64
Common mistakes that cost marks
- Writing the 3rd term as a + 3d. The nth term has (n − 1) differences: a + 2d.
- Treating the last term as the 49th term or as a + 50d. It is the 50th term, a + 49d.
- Finding d correctly but then using a + 29d for the 29th term (66).
How this can come in the exam
MCQ (1 mark)
In an AP, the 2nd term is 9 and the 6th term is 25. The common difference is
- 3
- 4
- 5
- 16
Show answer
(B) 4
(a + 5d) − (a + d) = 4d = 16 ⇒ d = 4.
Short answer (3 marks)
An AP has 30 terms. Its 4th term is 11 and its last term is 89. Find its 20th term.
Show answer
a + 3d = 11 and a + 29d = 89 (1 mark). Subtracting: 26d = 78 ⇒ d = 3, a = 2 (1 mark). 20th term = 2 + 19 × 3 = 59 (1 mark).Try one yourself
An AP has 40 terms. Its 5th term is 23 and its last term is 128. Find its 21st term.
Show answer
a + 4d = 23, a + 39d = 128 ⇒ 35d = 105, d = 3, a = 11. 21st term = 11 + 60 = 71.
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