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 15th term.

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Solution

Let a be the first term and d be the common difference of given AP.
Now, According to the question,
$a_{8} = \frac{1}{2} (a_{2})$ and $a_{11} = \frac{1}{3} (a_{4}) + 1$
$\Rightarrow 2 \times (a + 7 d) = a + d \Rightarrow a + 13 d = 0$ ...(1)
and $3 \times (a + 10 d) = (a + 3 d) + 3 \Rightarrow 2 a + 27 d = 3$ ...(2)
On applying $(2) - 2 \times (1)$ , we get
$d = 3$
Putting d=3 in (1), we get
$a + 13 \times 3 = 0 \Rightarrow a = - 39$
Now, $a_{15} = a + 14 d = - 39 + 14 \times 3 = 3$
$∴ a_{15} = 3$

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