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The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15^{th} term.

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#### Solution

Let a and d the first term and common difference of an AP, respectively.

Now, by given condition,

a_{8} = `1/2 a_2`

⇒ a + 7d = `1/2 (a + d)` ...[∵ a_{n} = a + (n – 1)d]

⇒ 2a + 14d = a + d

⇒ a + 13d = 0 ...(i)

And a_{11} = `1/3 a_4 + 1` ...[Given]

⇒ a + 10d = `1/3[a + 3d] + 1`

⇒ 3a + 30d = a + 3d + 3

⇒ 2a + 27d = 3

From equations (i) and (ii),

2(–13d) + 27d = 3

⇒ –26d + 27d = 3

⇒ d = 3

From equation (i),

a + 13(3) = 0

⇒ a = – 39

∴ a_{15} = a + 14d

= – 39 + 14(3)

= – 39 + 42

= 3

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