let the first term is [tex]a[/tex]
by the sum of first n term formula
[tex]\frac{20}{2}(2a+19d)=25a
\\20a+190d=25a15
\\190d=5a
\\a=38d[/tex]
substitute to sum of first 30 term
[tex]\frac{30}{2}(2a+29d)=\frac{30}{2}(2(38d)+29d)
\\=15(76d+29d)
\\=15(105d)
\\=1575d[/tex]
