Algebra

Progression & Series

Medium

On day one, there are $100$ particles in laboratory experiment. On day $n$ , where $n \geq 2$ , one out of every $n$ particles produces another particle. If the total number of particles in the laboratory experiment increases to $1000$ on day $m$ , then $m$ equals.

$19$

$16$

$17$

$18$

✅ Correct Option: 1

Day 1: 100 particles (given)

Day 2: On day $n$ , "1 out of every $n$ particles produces another particle"

$\therefore$ 1 out of every 2 particles produces another particle

New particles $= \frac{1}{2} \times 100 = 50$

Total particles $= 100 + 50 = 150$

Day 3:

New particles $=\frac{1}{3} \times 150 = 50$

Total particles $=150 + 50 = 200$

Day 4:

New particles $= \frac{1}{4} \times 200 = 50$

Total particles $= 200 + 50 = 250$

Notice something interesting? Each day we're adding exactly 50 particles!

This happens because:

Day 2: $\frac{1}{2} \times 100 = 50$

Day 3: $\frac{1}{3} \times 150 = 50$

Day 4: $\frac{1}{4} \times 200 = 50$

The pattern continues: $\frac{1}{n} \times (\text{particles on day } n-1) = 50$

Since we add 50 particles every day starting from day 2, our sequence is:

Day 1: 100

Day 2: 150

Day 3: 200

Day 4: 250

And so on...

This is an Arithmetic Progression with:

First term (a) = 100

Common difference (d) = 50

The general formula for the nth term of an AP is:

$a_n = a + (n-1)d$

In our case:

$\text{Particles on day n} = 100 + (n-1) \times 50$

Setting this equal to 1000:

$100 + (n-1) \times 50 = 1000$

$(n-1) \times 50 = 1000 - 100 = 900$

$n-1 = \frac{900}{50} = 18$

$n = 19$

On day 19, the laboratory will have exactly 1000 particles.

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On day one, there are $100$ particles in laboratory experiment. On day $n$ , where $n \geq 2$ , one out of every $n$ particles produces another particle. If the total number of particles in the laboratory experiment increases to $1000$ on day $m$ , then $m$ equals.