If the first term of AP is 4 and the sum of first five term is equal to one-fourth of the sum of the next five term, then find the 15th term?
Let $a$ denote the first term of the AP and let $d$ be the constant so that the $n$th term is $a+(n-1)d$. The sum of the first five terms is $5\cdot(a + a+4d)/2 = 5\cdot(a+2d) = 20 +10d$. The sum of the next five terms is $5\cdot((a+5d) + (a+9d))/2 = 5\cdot(a+7d) = 20 +35d$. The former is one-fourth of the latter, so $20+10d=(20+35d)/4$, which solves to $d=-12$. Therefore, the 15th term is $a+14d = 4-14\cdot 12=-164$.