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Explanation:

The last number is 3100.

Sum of the remaining numbers = S = 1 + 3 + 32 + … + 399   …(1)

This is a GP with a = 1; r = 3 and n = 100

∴ S = 1×(3100-1)3-1
 
∴ Required ratio = 31001×(3100-1)2 = 21-13100
 
∵ 1/3100 is approximately zero, hence it can be ignored.
 
∴ Ratio = 2
 
Hence, option (b).

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