Question

If $6$ times the sixth term of an arithmetic progression is equal to $9$ times the $9$th term, find the $15$th term.

Answer 1

$6(a+5d)=9(a+8d)$

or, $6a+30d=9a+72d$

or, $3a+42d=0$

or, $a+12d=0$

or, $a=-12d$

Now, $a_{15}=a+14d=-12d+14d=2d$