Different A.P.'s are constructed with the first term 100, the last term 199, and integral common dif

Q: Why can't we just pick any $d$ for the common difference?

That is a fantastic question! The problem explicitly states that the common difference $d$ must be an integer. Because our equation is $99 = (n-1)d$, this forces $d$ to be a divisor of $99$. We cannot simply choose any value; the structure of the arithmetic progression and the integer constraint restrict our choices to the factors of $99$.

Q: Why is the number of jumps $(n-1)$ and not $n$?

Think of it like a fence. If you have $n$ fence posts, there are only $n-1$ gaps between them. Similarly, in an arithmetic progression, to get from the first term to the $n$-th term, you take $n-1$ steps of size $d$. It is a fundamental property of discrete sequences that you must master!

Q: How do we handle the constraint $3 \le n \le 33$ effectively?

The best way is to transform the constraint to match our variable. Since we are working with $(n-1)$, subtract $1$ from all parts of the inequality $3 \le n \le 33$. This gives us $2 \le n-1 \le 32$. Now, you only need to look for divisors of $99$ that fall within this specific window. It turns the problem into a simple filtering exercise!

Question

Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the number of arithmetic progressions that start at $a = 100$ and end at $l = 199$, where the common difference $d$ is an integer. The number of terms $n$ is constrained by the condition $3 \le n \le 33$.

The general term formula for an arithmetic progression is given by:

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

Substituting our known values, we obtain:

$$199 = 100 + (n-1)d$$

### The Master Equation

By rearranging the terms, we isolate the total gap between the start and the finish:

$$199 - 100 = (n-1)d$$

$$99 = (n-1)d$$

This equation implies that the total gap of $99$ must be perfectly partitioned into $(n-1)$ jumps of integer size $d$.

### The Number Theory Constraint

Since $d$ must be an integer, $(n-1)$ must be a divisor of $99$. We identify the divisors of $99$ by finding all pairs of integers that multiply to $99$:

$$99 = 1 	imes 99$$
$$99 = 3 	imes 33$$
$$99 = 9 	imes 11$$

Thus, the set of possible values for $(n-1)$ is $\{1, 3, 9, 11, 33, 99\}$.

### The Filter of Reality

We must satisfy the constraint $3 \le n \le 33$. Subtracting $1$ from each part of this inequality, we define the valid range for the number of jumps:

$$2 \le n-1 \le 32$$

Applying this filter to our set of divisors $\{1, 3, 9, 11, 33, 99\}$, we find that only $3, 9,$ and $11$ fall within the range $[2, 32]$.

### Final Calculation

We now calculate the common difference $d$ for each valid case using the relation $d = \frac{99}{n-1}$:

1. For $n-1 = 3$, we have $d = \frac{99}{3} = 33$.
2. For $n-1 = 9$, we have $d = \frac{99}{9} = 11$.
3. For $n-1 = 11$, we have $d = \frac{99}{11} = 9$.

The possible values for the common difference $d$ are $33, 11,$ and $9$. The sum of these values is:

$$33 + 11 + 9 = 53$$

The final result is **53**.

Animated Solution for Mathematics - Sequence and Series: Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ________.

Enter Numerical Value:

Visualized Solution

Visualizing the A.P. Boundaries

The General Term Formula

Substituting Known Values

Finding the Total Gap

The Gap Equation

The Integer Constraint

Listing the Divisors

Constraint on Number of Terms

Adjusting the Constraint

Filtering Valid Divisors

Calculating Common Differences

Final Summation

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

Analyzing the Setup

The Master Equation

The Number Theory Constraint

The Filter of Reality

Final Calculation

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