Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ….
Answer: t10 = 48, t26 = 128.
Step-by-step solution
Idea: For an AP, tn = a + (n − 1)d. Here a = 3 and d = 5.
- a = 3, d = 8 − 3 = 5.½ mark
- t10 = 3 + (10 − 1) × 5 = 3 + 45 = 48.½ mark
- t26 = 3 + (26 − 1) × 5 = 3 + 125 = 128.1 mark
10th term = 48, 26th term = 128.
Check: tn = 5n − 2: 5 × 10 − 2 = 48 ✓, 5 × 26 − 2 = 128 ✓.
Answer to write in the exam
a = 3, d = 8 − 3 = 5
tn = a + (n − 1)d
t10 = 3 + 9 × 5 = 48
t26 = 3 + 25 × 5 = 128
∴ 10th term = 48, 26th term = 128
Common mistakes that cost marks
- Using 3 + 10 × 5 = 53 for the 10th term. Only 9 steps of 5 are added after the first term.
- Taking d = 3 (the first term) instead of 5.
- Arithmetic slip 25 × 5 = 120. It is 125.
How this can come in the exam
MCQ (1 mark)
The 15th term of the AP 3, 8, 13, 18, … is
- 73
- 75
- 78
- 70
Show answer
(A) 73
3 + 14 × 5 = 73.
Try one yourself
Find the 12th and 40th terms of the AP 7, 11, 15, 19, …
Show answer
a = 7, d = 4: 12th = 7 + 44 = 51; 40th = 7 + 156 = 163.
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