ARITHMETIC PROGRESSIONS: One of the endlessly alluring aspects of mathematics is that its thorniest paradoxes have a way of blooming into beautiful theories
Examples:
The fourth term of an AP is 0. Prove that its 25th term is triple its 11th term.
Ans: a_{4}= 0
⇒ a + 3d = 0
T.P a_{25}= 3 (a11)
⇒ a + 24d = 3 (a + 10d)
⇒ a + 24d = 3a + 30d
RHS sub a = - 3d
- 3d + 24d = 21d
LHS 3a + 30d
- 9d + 30d = 21d
LHS = RHS Hence proved