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

$a_{6}=a+5 d=0$

And

$a_{33}=a+32 d=a+5 d+27 d$

$=27 d \quad(\because a+5 d=0) \quad \ldots(1)$

3 times of 15 th term means

$3 a_{15}=3(a+14 d)=3 a+42 d$

$=3 a+15 d+27 d$

$=3(a+5 d)+27 d$

$=27 d$                    $\cdots \cdot(2)$

Thus, (1) = (2)
Hence, 33rd term is three times its 15th term.