A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 com

Q: Wait, why is the number of terms $n$ equal to 25? The problem says it took 8 more days, not 25.

That is a fantastic observation! The problem states the original plan was 17 days, and due to the crashes, it took 8 *more* days. So, the total time spent working is the original 17 days plus the additional 8 days. $17 + 8 = 25$. Since the systems were working (or crashing) every single day until the end, our sequence of work must span all 25 days.

Q: Why is the common difference $d$ negative? Shouldn't it be positive since the systems are crashing?

Think about what the variable represents: the number of *working* systems. On Day 1, we have $m$ systems. On Day 2, we have $m-4$. On Day 3, we have $m-8$. Since the number of working systems is decreasing, the change from one day to the next is a reduction. In an Arithmetic Progression, a reduction is represented by a negative common difference. Here, $d = -4$.

Q: Why can't we just use the average number of systems to solve this?

You *could* potentially use the average, but the Arithmetic Progression formula is much more robust and less prone to errors. The average of an A.P. is simply the average of the first and last terms. However, using the sum formula $S_n = \frac{n}{2} [2a + (n-1)d]$ is the standard, rigorous way to handle this. It ensures you don't miss any terms or miscalculate the total work done.

Question

A software company sets up $m$ number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of $m$ is equal to:

Options:

(A) 150

(B) 180

(C) 160

(D) 125

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are the project manager at a high-stakes software firm. You have a massive assignment, a deadline, and a team of computer systems ready to crunch the data. You have calculated that with $m$ systems, you can finish the job in exactly 17 days.

In physics and mathematics, when you encounter a problem involving work, the most powerful tool in your arsenal is the concept of **Invariance**. Think of the 'Total Work' as a fixed volume of water in a tank. Whether you drain it quickly with a wide pipe or slowly with a narrow one, the total amount of water remains the same.

In our scenario, the 'Total Work' is the product of the number of systems and the number of days. We define this as:

$$Total Work = m 	imes 17 = 17m$$

This $17m$ is our North Star. No matter how many systems crash, the total effort required to complete the assignment is fixed at $17m$ system-days. This is the bedrock upon which we will build our solution.

### The Pattern of Chaos

Now, let us look at the chaos. On the first day, all $m$ systems are humming along. But at the start of the second day, disaster strikes: 4 systems crash. Now we have $m-4$. On the third day, 4 more crash, leaving us with $m-8$.

Do you see the pattern? It is not random; it is a sequence. Specifically, it is an **Arithmetic Progression (A.P.)**. In an A.P., we have a first term, $a$, and a common difference, $d$.

Here, our first term is $a = m$. Our common difference is $d = -4$ (because the number of systems is decreasing by 4 each day).

### The Time Dilation

We know the original plan was 17 days. But the problem tells us it took 8 *more* days to finish. If the work took 8 more days, the total duration is simply:

$$n = 17 + 8 = 25 	ext{ days}$$

Our A.P. does not stop at 17; it continues for 25 days. We are now looking for the sum of the first 25 terms of this sequence.

### The Mathematical Bridge

We have the total work ($17m$) and we have the actual work done (the sum of the A.P. over 25 days). Since the work is invariant, we can equate them.

Recall the sum formula for an A.P.:

$$S_n = \frac{n}{2} [2a + (n-1)d]$$

Substituting our known values ($n=25$, $a=m$, $d=-4$):

$$Actual Work = \frac{25}{2} [2m + (25-1)(-4)]$$

Now, we set the Planned Work equal to the Actual Work:

$$17m = \frac{25}{2} [2m + 24(-4)]$$

### The Algebraic Resolution

This is the moment where the complexity dissolves. Let us simplify the expression inside the brackets. Since $24 	imes -4 = -96$, we have:

$$17m = \frac{25}{2} [2m - 96]$$

Notice the beauty of the numbers here. We can factor out a 2 from the bracket, which perfectly cancels the 2 in the denominator:

$$17m = 25 [\frac{2m}{2} - \frac{96}{2}]$$

$$17m = 25(m - 48)$$

Now, it is just simple algebra. Expand the right side:

$$17m = 25m - 1200$$

Rearrange to isolate $m$:

$$25m - 17m = 1200$$

$$8m = 1200$$

$$m = \frac{1200}{8}$$

$$m = 150$$

### Conclusion: The Elegance of Logic

And there we have it. The initial number of systems was **150**.

What makes this problem so satisfying is not the final number, but the journey. We took a chaotic real-world scenario—systems crashing, deadlines slipping—and we tamed it with the rigid, beautiful structure of an Arithmetic Progression.

Remember, in JEE Advanced, the math is rarely just about calculation. It is about modeling. It is about finding the invariant truth in a sea of variables. You have just mastered that skill. Keep that confidence, and carry it into your next challenge!

Select Answer:

Visualized Solution

Defining Total Work

Analyzing the System Crashes

Calculating Total Actual Days

Applying the Sum of A.P. Formula

Equating the Total Work

Simplifying the Equation

Expanding and Rearranging

Final Calculation

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Pattern of Chaos

The Time Dilation

The Mathematical Bridge

The Algebraic Resolution

Conclusion

Similar Questions

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + .....$ is $(102) m$ , then $m$ is equal to:

A man saves Rs $200$ in each of the first three months of his service. In each of the subsequent months his saving increases by Rs $40$ more than the saving of immediately previous month. His total saving from the start of service will be Rs $11040$ after

If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + \dots$ is $(102) m$ , then m is equal to :

Let $a_{1}, a_{2}, a_{3}, \dots, a_{49}$ be in A.P. such that $\sum_{k = 0}^{12} a_{4 k + 1} = 416$ and $a_{9} + a_{43} = 66$ . If $a_{1}^{2} + a_{2}^{2} + \dots + a_{17}^{2} = 140 m$ , then $m$ is equal to :

Suppose $a_{1}, a_{2}, \dots, a_{n}, \dots$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $5 : 17$ and $110 < a_{15} < 120$ , then the sum of the first ten terms of the progression is equal to -

Let $T_{r}$ be the rth term of an A.P. If for some m, $T_{m} = \frac{1}{25}$ and $T_{25} = \frac{1}{20}$ , then $5 m \sum_{r = m}^{2 m} T_{r}$ is equal to Note: $20 \sum_{r = 1}^{25} T_{r} = 13$ is given condition in Hindi version.

A person is to count $4500$ currency notes. Let $a_{n}$ denote the number of notes he counts in the $n$ th minute. If $a_{1} = a_{2} = \dots = a_{10} = 150$ and $a_{10}, a_{11}, \dots$ are in an AP with common difference $- 2$ , then the time taken by him to count all notes is

If $a_{1}, a_{2}, a_{3}, \dots$ are in A.P. such that $a_{1} + a_{7} + a_{16} = 40$ , then the sum of the first 15 terms of this A.P. is :

Let $a_{1}, a_{2}, \dots, a_{n}$ be in A.P. If $a_{5} = 2 a_{7}$ and $a_{11} = 18$ , then $12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots + \frac{1}{a _{17} + a _{18}})$ is equal to

Let $a_{1}, a_{2}, \dots, a_{21}$ be an $A P$ such that $\sum_{n = 1}^{20} \frac{1}{a _{n} a _{n + 1}} = \frac{4}{9}$ . If the sum of this $A P$ is $189$ , then $a_{6} a_{16}$ is equal to :