Let a1,a2,…,a30 be an A.P., S=∑i=130 ai and T=∑i=115 a(2i−1). If a5=27 and S−2T=75 then a10 is
S=a1+a2+...+a30
=152a1+29dT=a1+a3+…+a29T=1522a1+14×2d−−−12T=152a1+28d−−−2S−2T=15d=75d=5a10=a5+5d=27+25=52