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Exercise 5.3 · Q12

Q.In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particles gets doubled each day, being 55 particles on the first day. Find the day when the infectious virus particles just grow over 1,50,0001{,}50{,}000 units.

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The virus count is a GP doubling daily from 5; find the smallest day index where the count exceeds 150000.

Step 1. Set up the GP. Day nn count =5×2n−1=5\times2^{n-1} (day 1: 55 particles, doubling each subsequent day).

Step 2. Set up the inequality. 5×2n−1>150000⇒2n−1>300005\times2^{n-1} > 150000 \Rightarrow 2^{n-1} > 30000.

Step 3. Find the smallest power of 2 exceeding 30000. 214=163842^{14}=16384 (too small), 215=327682^{15}=32768 (>30000>30000). So n−1=15⇒n=16n-1=15 \Rightarrow n=16. …

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