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Question
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
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Solution
Let a be the first term and d be the common difference of the AP. Then,
as = 0 [ an = 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 38th term of the AP id triple its 18th term.
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