If the first term of A.P is 2 and the sum of first five terms is equal to one fourth of the sum of the next five terms then the sum of first 30 terms is

${\mathrm{T}}_1+{\mathrm{T}}_2+{\mathrm{T}}_3+{\mathrm{T}}_4+{\mathrm{T}}_5=\frac{1}{4}\left[{\mathrm{T}}_6+{\mathrm{T}}_7+{\mathrm{T}}_8+{\mathrm{T}}_9+{\mathrm{T}}_{10}\right]$

$\begin{array}{l}\frac{5}{2}\left({\mathrm{T}}_1+{\mathrm{T}}_{\underset{ }{5}}\right)=\frac{1}{4}\cdot \frac{5}{2}\left({\mathrm{T}}_6+{\mathrm{T}}_{10}\right)\\ \to 2\mathrm{a}+4\mathrm{d}=\frac{1}{4}(2\mathrm{a}+14\mathrm{d}) \left( \sin ce t_n=a+\left(n-1\right)d\right)\\ \to \mathrm{d}=-3\mathrm{a}=-3\left(2\right)=-6\\ \left( \sin ce a=2\right)\\ {\mathrm{S}}_{30}=\frac{30}{2}\left[2\left(2\right)+(30-1)(-6)\right]=15\mathrm{x}-170=-2550\\ \left(\sin ce s_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\right)\end{array}$