An AP has common difference 3, and its 15th term is 5 more than twice its 5th term. Find the first term.

$a_{15}=a+42$ and $a_5=a+12$. From $a+42=2(a+12)+5$ we get $a+42=2a+29$, so $a=13$. (Check: $a_5=25,a_{15}=55$ and 2*25 + 5 = 55.) Dropping the '+5' gives 18, using 15d and 5d instead of 14d and 4d gives 10, and 25 is the 5th term rather than the first.