1. Interpret the data

He counts notes minute-wise, where $a_n$ is the number of notes counted in the $n^{\text{th}}$ minute.

Given:

• $a_1=a_2=\cdots=a_{10}=150$
• From $a_{10},a_{11},a_{12},\dots$ the terms are in an AP with common difference $-2$

So:

• In the first $10$ minutes, he counts $150$ notes each minute.
• After that, the counts are: $150,148,146,144,\dots$ starting from the $10^{\text{th}}$ minute onward.

2. Count notes in the first 10 minutes

$S_{10}=10\times 150=1500$

So, after $10$ minutes, remaining notes are:

$4500-1500=3000$

3. Now count the remaining notes using the AP part

From the $11^{\text{th}}$ minute onward, the numbers counted per minute form:

$148,146,144,\dots$

Let he takes $m$ more minutes after the first $10$ minutes. Then these $m$ terms form an AP with:

• first term $a=148$
• common difference $d=-2$

Their sum is:

$S_m=\frac{m}{2}\left[2(148)+(m-1)(-2)\right]$

$S_m=\frac{m}{2}\left[296-2m+2\right]$

$S_m=\frac{m}{2}(298-2m)=m(149-m)$

This must equal $3000$:

$m(149-m)=3000$

$149m-m^2=3000$

$m^2-149m+3000=0$

4. Solve the quadratic

$m^2-149m+3000=0$

Factorizing:

$3000=24\times 125, \quad 24+125=149$

So,

$m^2-149m+3000=(m-24)(m-125)=0$

Hence,

$m=24 \quad \text{or} \quad m=125$

5. Choose the meaningful value

If $m=125$, then the last term would be:

$148+(125-1)(-2)=148-248=-100$

This is impossible, since number of notes counted in a minute cannot be negative.

So only

$m=24$

is valid.

6. Total time

Total time taken:

$10+24=34 \text{ minutes}$

7. Check with options

Option A: $34$ minutes

So the correct answer is:

$\boxed{\text{A: 34 minutes}}$