First slide
Arithmetic progression
Question

 Let a1,a2,,a30 be an A.P., S=i=130ai and T=i=115a(2i1). If a5=27 and S2T=75 then a10 is

Moderate
Solution

S=a1+a2+...+a30

=152a1+29dT=a1+a3++a29T=1522a1+14×2d12T=152a1+28d2S2T=15d=75d=5a10=a5+5d=27+25=52

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