Since t7 – t5 = 12, we have 2d = 12, giving d = 6. With third term a + 2d = 16, substituting d gives a = 4. The required AP is 4, 10, 16, 22, ….
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Let the first term be a and common difference be d.
Given:
t7 – t5 = 12
(a + 6d) – (a + 4d) = 12
2d = 12
d = 6.
Also, the third term is 16:
t3 = a + 2d = 16
a + 2 x 6 = 16
a + 12 = 16
a = 4.
With first term a = 4 and common difference d = 6, the terms are:
First term = 4
Second term = 4 + 6 = 10
Third term = 10 + 6 = 16
Fourth term = 16 + 6 = 22.
Thus, the arithmetic progression is 4, 10, 16, 22, ….
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/