Let a1, a2,....an be a given A.P. whose common difference is an integer and Sn = a1 + a2 + .... +an

Q: Wait, why do we reject the other divisors of 299? Shouldn't there be multiple solutions?

That is a fantastic observation! Mathematically, $299$ has divisors $1, 13, 23,$ and $299$. However, in JEE problems, we are often constrained by the physical or logical boundaries of the sequence. Here, the condition $15 < n < 50$ acts as a filter. By subtracting $1$ from the inequality, we get $14 < n-1 < 49$. Only the divisor $23$ falls within this specific range, which is why we discard the others. Always check your constraints!

Q: Why is it so important that the common difference $d$ is an integer?

Great question! If $d$ were not restricted to an integer, there would be infinitely many pairs of $(n, d)$ that satisfy the equation $299 = (n-1)d$. By specifying that $d \in \mathbb{Z}$, the problem transforms from a general algebraic equation into a Diophantine equation, limiting the possible values of $n$ and $d$ to a finite set. This is a classic technique used to make problems solvable.

Q: Can I use the sum formula $S_n = \frac{n}{2}(2a_1 + (n-1)d)$ instead of $S_n = \frac{n}{2}(a_1 + a_n)$?

Absolutely! Both formulas are mathematically equivalent. However, using $S_n = \frac{n}{2}(a_1 + a_n)$ is often faster and less prone to calculation errors because you are using the values already provided or calculated ($a_1=1$ and $a_n=300$). Always look for the most efficient path to the solution to save precious time during the exam.

Question

Let $a_1, a_2,....a_n$ be a given A.P. whose common difference is an integer and $S_n = a_1 + a_2 + .... +a_n$ If $a_1 = 1, a_n = 300$ and $15 < n < 50$, then the ordered pair $(S_{n-4}, a_{n-4})$ is equal to

Options:

(A) (2490,249)

(B) (2480, 249)

(C) (2490, 248)

(D) (2480, 248)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the start of a long, perfectly straight path. You are at the first marker, $a_1 = 1$. Your goal is to reach the $n$-th marker, $a_n = 300$.

Between these two points, you are taking consistent, equal steps of size $d$. This is the essence of an Arithmetic Progression (AP).

The relationship between our start, our end, and our steps is governed by the elegant formula:

$$a_n = a_1 + (n-1)d$$

This equation is our bridge. It tells us that to get from the first term to the $n$-th term, we must traverse exactly $n-1$ intervals of length $d$.

### The Integer Constraint and the Number Theory Trap

When we substitute our known values into the bridge formula, we get $300 = 1 + (n-1)d$, which simplifies beautifully to:

$$299 = (n-1)d$$

Now, here is where the problem tests your mathematical intuition. We have one equation and two unknowns, $n$ and $d$, but we are given a secret weapon: $d$ is an integer.

This means that $(n-1)$ must be a divisor of $299$. To unlock this, we perform a prime factorization of $299$. Since $299$ is not divisible by $2$ or $3$, we test higher primes and discover that:

$$299 = 13 	imes 23$$

Thus, the possible values for $(n-1)$ are $1, 13, 23,$ and $299$.

### Navigating the Constraints

We are not done yet! We have a range constraint: $15 < n < 50$. If we subtract $1$ from all parts of this inequality, we get:

$$14 < n-1 < 49$$

Looking at our list of divisors—$1, 13, 23, 299$—only one number sits comfortably in that window: $23$. This is the moment of clarity.

We now know that $n-1 = 23$, which implies $n = 24$. With $n$ in hand, finding $d$ is trivial:

$$d = \frac{299}{23} = 13$$

We have successfully decoded the sequence!

### The Final Calculation

The question asks for the ordered pair $(S_{n-4}, a_{n-4})$. Since $n=24$, we are looking for $(S_{20}, a_{20})$.

First, let's find the twentieth term, $a_{20}$. Using our formula $a_k = a_1 + (k-1)d$, we calculate:

$$a_{20} = 1 + 19(13) = 1 + 247 = 248$$

Now, for the sum $S_{20}$. We use the efficient sum formula:

$$S_k = \frac{k}{2}(a_1 + a_k)$$

Substituting our values, we get:

$$S_{20} = \frac{20}{2}(1 + 248) = 10(249) = 2490$$

The final result is the ordered pair **$(2490, 248)$**. You have navigated the constraints, applied the number theory, and executed the calculation with precision.

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, .... a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2} + .... + a_{n}$ If $a_{1} = 1, a_{n} = 300$ and $15 < n < 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to

Select Answer:

Visualized Solution

Given Parameters

General Term of A.P.

Substitute Values

Integer Constraint

Factorize $299$

Range Constraint on $n$

Calculate $n$ and $d$

Target Terms

Calculate $a_{20}$

Sum Formula for $S_{20}$

Final Conclusion

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

Analyzing the Setup

The Integer Constraint and the Number Theory Trap

Navigating the Constraints

The Final Calculation

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