Let a1,a2…a3n be an arithmetic progression with a1=3 and a2=7. If a1+a2+…+a3n=1830, then what is the smallest positive integer m such that m(a1+a2+…+an)>1830?
9
a1=3a2=7
common difference =d=7−3=4a3n=a1+(3n−1)×4=3+12n−4=12n–13+7+11+…+(12n−1)=18303n[6+(3n−1)×4]2=18303n[2+12n]=3660n[1+6n]=6106n2+n−610=0n=10
Therefore, a10=3+9×4=39a1+a2+…+a10=3+7+…+39=5×42=210m×210>1830m>8
Hence smallest integral value of m = 9.