Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of

Q: Wait, why do we reject non-integer values for $d$?

The problem explicitly states that all terms of the arithmetic progression are natural numbers. Since $a_n = a + (n-1)d$, if $d$ were not an integer, the terms would not remain integers for all $n$. Thus, $d$ must be an integer.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure mathematics problem involving sequences and series. While physics problems often use energy conservation, here we rely on the algebraic properties of arithmetic progressions. Stick to the definitions of $S_n$ and $a_n$!

Q: Is it possible for $a$ to be negative?

No. The problem states that all terms are natural numbers. Natural numbers are defined as positive integers ($\{1, 2, 3, \dots\}$). Therefore, $a$ must be a positive integer.

Question

Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6 : 11$ and the seventh term lies in between $130$ and $140$, then the common difference of this A.P. is \dots.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the start of a long, perfectly paved path. You are about to take steps of a fixed length, $d$, starting from an initial position, $a$. This is the essence of an Arithmetic Progression (A.P.).

Every term is a predictable, rhythmic addition of the common difference to the previous term. In this problem, we are told that every single term in this sequence is a natural number.

This is our anchor, our guiding light. It means that both our starting point $a$ and our step size $d$ must be positive integers.

### Decoding the Ratio

We are given a fascinating clue: the ratio of the sum of the first seven terms, $S_7$, to the sum of the first eleven terms, $S_{11}$, is exactly $6:11$. Let's invoke our trusty sum formula:

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

For $n=7$, we have $S_7 = \frac{7}{2}[2a + 6d]$. For $n=11$, we have $S_{11} = \frac{11}{2}[2a + 10d]$.

When we set up the ratio $\frac{S_7}{S_{11}} = \frac{6}{11}$, the $\frac{1}{2}$ terms in the numerator and denominator cancel out. By simplifying the expression, we get:

$$\frac{7(a+3d)}{11(a+5d)} = \frac{6}{11}$$

Notice how the $11$ in the denominator of the left side and the $11$ on the right side cancel out. We are left with a beautifully simple linear equation: $7(a+3d) = 6(a+5d)$.

Expanding this, we get $7a + 21d = 6a + 30d$. Subtracting $6a$ from both sides and $21d$ from both sides, we arrive at the elegant result:

$$a = 9d$$

### The Constraint of the Seventh Term

Now, we turn to our second clue. The seventh term, $a_7$, is trapped between $130$ and $140$. Using the general term formula $a_n = a + (n-1)d$, we know that $a_7 = a + 6d$.

Substituting our breakthrough relation $a = 9d$ into this, we get $a_7 = 9d + 6d = 15d$. So, our inequality becomes:

$$130 < 15d < 140$$

To find $d$, we simply divide the entire inequality by $15$:

$$\frac{130}{15} < d < \frac{140}{15}$$

Calculating these bounds, we find $8.66 < d < 9.33$.

### The Final Step

Here is where the initial condition saves us. We know $d$ must be an integer because all terms of the A.P. are natural numbers.

Looking at our range, there is only one integer that fits perfectly between $8.66$ and $9.33$. That integer is $9$.

And there it is! The common difference of our A.P. is **$9$**. It is a testament to how constraints in mathematics act as filters that guide us to the unique, elegant solution.

Enter Numerical Value:

Visualized Solution

Define the A.P. and Given Conditions

Apply the Sum Formula $S_{n}$

Substitute into the Ratio

Simplify the Ratio Equation

Establish Relation Between $a$ and $d$

Express the Seventh Term $a_{7}$

Apply the Inequality Constraint

Solve for $d$

Conclusion and Key Takeaway

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

Analyzing the Setup

Decoding the Ratio

The Constraint of the Seventh Term

The Final Step

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