Given a + 2d = 12 and a + 49d = 106, solving these simultaneous equations yields first term a = 8 and common difference d = 2. Therefore, the 29th term evaluates to 8 + 28 x 2 = 64.
An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Share
Given an AP of 50 terms:
3rd term: a + 2d = 12
50th term: a + 49d = 106
Subtracting the first equation from the second:
(a + 49d) – (a + 2d) = 106 – 12
47d = 94, which gives d = 2.
Substitute d = 2 into the first equation:
a + 2 x 2 = 12
a = 12 – 4 = 8.
Now, finding the 29th term:
t29 = a + 28d = 8 + 28 x 2 = 8 + 56 = 64.
Hence, the 29th term is 64.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/