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

T1+T2+T3+T4+T5=14T6+T7+T8+T9+T1052T1+T5 =14⋅52T6+T10→2a+4d=14(2a+14d)    since tn=a+n-1d→d=−3a=−32=−6          since a=2S30=30222+(30−1)(−6)=15x−170=−2550     since sn=n22a+n-1d