Arithmetic progressions, Mathematics

ARITHMETIC PROGRESSIONS: One  of the  endlessly alluring  aspects  of mathematics  is  that its thorniest  paradoxes have  a  way  of blooming  into  beautiful  theories


The fourth term of an AP is 0. Prove that its 25th term is triple its 11th term.

Ans:   a4= 0

⇒ a + 3d = 0

T.P       a25= 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


Posted Date: 4/8/2013 5:14:49 AM | Location : United States

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