Question

The 8^th term of an AP is zero. Prove that its 38^th term is triple its 18^th term.  

Answer 1

Let a be the first term and d be the common difference of the AP. Then,

a_s  = 0             [ a_n = a + (n-1) d]

⇒ a + (8-1 ) d = 0

⇒ a+ 7d =0

⇒ a=-7d                 ..................(1)

Now 

`(a_38)/(a_18) = (a + (38-1)d)/(a+ (18-1)d)`

 ⇒`(a_38)/(a_18) = (-7d + 37 d)/(-7d + 17d ) `           [ From (1)]

⇒`(a_38)/(a_18) = (30d)/(10d) = 3`

⇒` a_38 = 3 xx a_18`

Hence, the 38^th term of the AP id triple its 18^th term.