The minimum number of terms of the series  1+ 3 + 9 + 27 + … so that the sum may exceed 1000, is

Let, the sum of n terms exceeds 1000.Then 11−3n1−3>1000; or 3n−12>1000 or 3n>2001; but 36=729 and 37=2187; ∴n>6;  but n is a positive integer∴n=7,8,9,…∴ The minimum number of terms = 7.