In the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, . . . . . , where $$n$$ consecutive terms have the value $$n,$$ the $$1025^{th}$$  term is

Solution :
Let the $$1025^{th}$$  term = $$2^n.$$ Then
$$\eqalign{ & 1 + 2 + 4 + 8 + ..... + {2^{n - 1}} < 1025 \leqslant 1 + 2 + 4 + 8 + ..... + {2^n} \cr & \therefore \,\,{2^n} - 1 < 1025 < {2^{n + 1}} - 1\,\,{\text{or, }}{2^n} < 1026 < {2^{n + 1}} \cr & \Rightarrow \,\,n = 10. \cr} $$