In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms, then $\text{'}d\text{'}=$

$a=2\text{\hspace{0.17em}and\hspace{0.17em}}2,2+d,2+2d,2+3d,2+4d$ are first five terms

Sum of 1^st 5 terms $=10+10d$

Sum of next 5 terms $=10+35d$

$=2+5d+2+6d+2+7d+2+8d+2+9d$ are next five terms

As per problems

$10+10d=\frac{1}{4}(10+35d)$

$40+40d=10+35d$

$d=-6$.