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Question
The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.
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Solution
Let a be the first term and d be the common difference of the AP. Then,
`a_19 = 3a_6` (Given)
⇒ a + 18d = 3 (a+ 5d) [ an = a + (n-1) d]
⇒ a + 18d = 3a + 15d
⇒ 3a - a = 18d - 15d
⇒ 2a = 3d ................(1)
Also,
a9 = 19 (Given)
⇒ a +8d = 19 ..............(2)
From (1) and (2), we get
`(3d)/2 + 8d = 19`
`⇒ (3d +16d)/2 = 19`
⇒ 19d =38
⇒ d =2
Putting d = 2 in (1), we get
2a = 3 × 2=6
⇒ a = 3
So,
a2 = a + d = 3+2 = 5
a3 = a +2d = 3+2 × 2=7
Hence, the AP is 3,5,7,9,........
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