If 8 times the 8th term of an AP is equal to 15 times the 15th term of that AP, then find the 23rd term of the same AP.

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Solution

Given,
$8 a_{8} = 15 a_{15}$
$\Rightarrow$ 8[a + (8 - 1)d] = 15[a + (15 - 1)d]
$\Rightarrow$ 8(a + 7d) = 15 (a + 14d)
$\Rightarrow$ 8a + 56d = 15a + 210d
$\Rightarrow$ -7a - 154d = 0
$\Rightarrow$ 7a + 154d = 0
$\Rightarrow$ a + 22d = 0
$\Rightarrow$ a + (23 - 1)d = 0

$∴$ $a_{23} = 0$

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