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Let the sequence have 1000 zeros .In step1,to every position in the sequence we add 2.In step 2 to every even position we add 2.In step 3 to every position which is multiple of 3 we add 2.This is continued to 1000th step.After 1000th step what will be the value in the 600th position?                                       a]48                                                                                           b]24c]64d124

krishna priya
28 Points
2 years ago
48
Step-by-step explanation:
Let the sequence be a₁,a₂,a₃,a₄,a₅,.........a₁₀₀₀.
Step 1: Every position in the sequence is increased by 2.
Step 2: Every even position in the sequence is increased by 2.
Step 3: Every  3rd term in the sequence is increased by 2.
........
Analysis:
a₁ is added by 2 only in step 1
a₂ is added by 2 in  steps 1 and 2
a₃ is added by 2 in steps 1 and 3
............
Clearly it is evident that any term aₓ is added by 2 in step n ,
if and only if n is the factor of x.
Now, if we factorize 600 we get , 600 =2³×3×5².
Thus total number of factors of 600 are (1+3)×(1+1)×(1+2) = 24.
So,  during each of these 24 steps a₆₀₀ will be added by 2.
Thus a₆₀₀ would be added by 2  24 times , hence a₆₀₀ would be
increased by 24×2 = 48.