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

Given, $a_{6} = 0 \Rightarrow a + 5 d = 0 \Rightarrow a = - 5 d$

Now, $a_{15} = a + 14 d = - 5 d + 14 d = 9 d$

and $a_{33} = a + 32 d = - 5 d + 32 d = 27 d = 3 \times 9 d = 3 a_{15}$

Hence, proved.

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