CAT 2023 Slot 1QAAlgebra • Progression & SeriesModerate
A lab experiment measures the number of organisms at 8am every day. Starting with 2organisms on the first day, the number of organisms on any day is equal to 3more than twice the number on the previous day. If the number of organisms on nth day exceeds one million, then the lowest possible value of n is
TITA Answer:
⌨️ Shortcuts:[1-4 / A-D] Select[Enter] Check[B] Star[S] Solution
Official Correct Answer: 19. The problem says that the experiment measures the number of organisms at 8am every day. It starts with 2 organisms on the first day. Each subsequent day, the number of organisms is equal to 3 more than twice the number on the previous day. We find the smallest day number, n, where the number of organisms exceeds one million. The correct answer is given as 19, but I need to figure out how to get there step by step.
Okay, let's break this down. It sounds like a sequence where each term is defined based on the previous one. That makes me think of geometric progressions or maybe recurrence relations. Let me recall the formula for a geometric progression. It's usually something like a_n = a_1 * r^(n-1), where a_1 is the first term and r is the common ratio. But wait, in this case, each term is twice the previous term plus 3. So it's not a simple geometric progression because of that extra 3. Hmm, maybe it's a linear recurrence relation.
We have the recurrence relation. If I let a_n be the number of organisms on day n, then according to the problem, a_n = 2*a_{n-1} + 3. And we know that a_1 = 2. So this is a linear nonhomogeneous recurrence relation. these can be solved by finding the homogeneous solution and a particular solution.
First, let's solve the homogeneous part. The homogeneous equation would be a_n - 2*a_{n-1} = 0. The characteristic equation is r - 2 = 0, so r = 2. Therefore, the homogeneous solution is a_n^h = C*2^n, where C is a constant.
Next, we need a particular solution. Since the nonhomogeneous term is a constant (3), we can assume a particular solution is a constant, say A. Plugging this into the recurrence relation: A = 2*A + 3. Solving for A: A - 2*A = 3 => -A = 3 => A = -3.
So the general solution is the sum of the homogeneous and particular solutions: a_n = C*2^n - 3. Now, we can use the initial condition to find C. On day 1, a_1 = 2. Plugging n=1 into the general solution: 2 = C*2^1 - 3 => 2 = 2C - 3 => 2C = 5 => C = 5/2.
Therefore, the explicit formula for the number of organisms on day n is a_n = (5/2)*2^n - 3. Simplifying that, (5/2)*2^n is 5*2^{n-1}, so a_n = 5*2^{n-1} - 3.
Now, we need to find the smallest n such that a_n > 1,000,000. So let's set up the inequality: 5*2^{n-1} - 3 > 1,000,000. Adding 3 to both sides: 5*2^{n-1} > 1,000,003. Dividing both sides by 5: 2^{n-1} > 1,000,003 / 5. Calculate that: 1,000,003 divided by 5 is 200,000.6. So 2^{n-1} > 200,000.6.
Now, I need to solve for n. Taking the logarithm base 2 of both sides: n - 1 > log2(200,000.6). Let me approximate log2(200,000.6). I know that 2^17 = 131,072 and 2^18 = 262,144. So 200,000.6 is between 2^17 and 2^18. , 2^17 is 131,072, which is less than 200,000.6, and 2^18 is 262,144, which is more. So log2(200,000.6) is between 17 and 18. To get a better estimate, Computing how much more than 2^17 is 200,000.6.
200,000.6 - 131,072 = 68,928.6. So the difference is about 68,928.6. The total range between 2^17 and 2^18 is 262,144 - 131,072 = 131,072. So the fraction is 68,928.6 / 131,072 ≈ 0.525. Therefore, log2(200,000.6) ≈ 17 + 0.525 ≈ 17.525.
So n - 1 > 17.525 => n > 18.525. Since n must be an integer, the smallest n is 19. Therefore, the lowest possible value of n is 19.
Starting with a_n = 5*2^{n-1} - 3. For n=19, a_19 = 5*2^{18} - 3. 2^10 is 1024, so 2^18 is 262,144. So 5*262,144 = 1,310,720. Subtract 3, we get 1,310,717, which is indeed more than one million. What about n=18? a_18 = 5*2^{17} - 3 = 5*131,072 - 3 = 655,360 - 3 = 655,357, which is less than one million. So yes, n=19 is the correct answer.
I think that makes sense. The number of organisms grows exponentially because each day it's doubling and adding a constant. So it doesn't take too many days to reach over a million. Using the formula we derived, we can quickly find the exact day when the count exceeds one million without