If the sixth term of an AP is zero then show that its 33rd term is three times its 15th term.

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Solution

$a_{6} = a + 5 d = 0 And a_{33} = a + 32 d = a + 5 d + 27 d = 27 d (∵ a + 5 d = 0) . . . . . (1) 3 times of 15 th term means 3 a_{15} = 3 (a + 14 d) = 3 a + 42 d = 3 a + 15 d + 27 d = 3 (a + 5 d) + 27 d = 27 d . . . . . (2)$
Thus, (1) = (2)
Hence, 33rd term is three times its 15th term.

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