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Question
The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
Solution
Let a be the first term and d be the common difference of the AP. Then,
`a_24 = 2a_10` (Given)
⇒ a + 23d = 2(a +9d) [an = a + (n-1) d]
⇒ a+ 23d = 2a - 18d
⇒ 2a - a = 23d -18d
⇒ a = 5d ..............(1)
Now ,
`(a_72)/(a_15) = (a+ 71d)/(a+14d) `
⇒ ` (a_72)/(a_15) = (5d + 71d)/(5d+14d)` [From(1)]
⇒ `(a_72)/(a_15) = (76d)/(19d)=4`
⇒ `a_72 = 4 xx a_15`
Hence, the 72nd term of the AP is 4 times its 15th term.
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