ICSE & CBSE Board Question Based on Arithmetic Progression of RS Aggarwal

A term of AP is given. you have to prove that the another *n*th term is Triple than its before *n*th term

This is the Question Number 18 Of Exercise 11 A of RS Aggarwal Solution.

Deepak BoraNewbie

# The 4th term of an AP is Zero. Prove that its 25th term is triple its 11th term.

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Given a^4 = 0

That is (a + 3d) = 0

⇒ a = – 3d ……….. (1)

nth term of AP is given by an = a + (n – 1)d

a^11 = a + 10d = – 3d + 10d = 7d [From (1)]

a^25 = a+ 24d = – 3d + 24d = 21d [From (1)]

= 3 x 7d

Hence a^25 = 3 x a^11