Suppose a₁, a₂, ..., aₙ, ... be an arithmetic progression of natural numbers. If the ratio of the su

Q: Wait, why do we reject the negative root or other values for $a$?

The problem restricts $a$ and $d$ to be natural numbers ($a, d \in \mathbb{N}$). When we solve the inequality $110 < 57a < 120$, we are looking for an integer $a$. Since $57 \times 1 = 57$ (too small) and $57 \times 3 = 171$ (too large), $a=2$ is the only natural number that satisfies the condition.

Q: Why can't we just solve for $a$ and $d$ directly from the ratio?

The ratio $\frac{S_5}{S_9} = \frac{5}{17}$ provides a linear relationship between $a$ and $d$, but it does not fix their absolute values. It only tells us the ratio of the growth of the sequence. We need the second piece of information (the inequality) to anchor the sequence to specific values.

Q: Is there a faster way to find $S_{10}$ once we have $a$ and $d$?

Once you have $a=2$ and $d=8$, you could list the terms, but using the sum formula $S_n = \frac{n}{2} [2a + (n-1)d]$ is the most robust and error-proof method. It is the standard tool for a reason!

Question

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 -

Options:

(A) 290

(B) 380

(C) 460

(D) 510

Sigma Percentile · Accepted Answer

### The Beauty of Arithmetic Progressions

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that, at first glance, might seem like a standard algebra exercise, but is actually a beautiful dance between ratios and integer constraints.

Let us embark on this journey together.

### Phase 1: Defining the Landscape

Every arithmetic progression (A.P.) is defined by two fundamental pillars: the first term, which we call $a$, and the common difference, $d$.

The problem gives us a vital clue right at the start: $a, d \in \mathbb{N}$. This means our sequence is built entirely from positive integers.

This is not just a definition; it is a constraint that will act as our 'lock' later on. When you see 'natural numbers' in a JEE problem, treat it as a massive hint that you will eventually be testing integer values.

### Phase 2: The Algebraic Dance

We are given the ratio of the sum of the first five terms to the sum of the first nine terms: $\frac{S_5}{S_9} = \frac{5}{17}$. Recall our trusty sum formula: $S_n = \frac{n}{2} [2a + (n-1)d]$.

Let us write this out for $n=5$ and $n=9$:

$$S_5 = \frac{5}{2} [2a + 4d] = 5(a + 2d)$$

$$S_9 = \frac{9}{2} [2a + 8d] = 9(a + 4d)$$

Now, we set up our ratio:

$$\frac{5(a + 2d)}{9(a + 4d)} = \frac{5}{17}$$

Notice the elegance here? The factor of $5$ on both sides cancels out immediately, leaving us with:

$$\frac{a + 2d}{9(a + 4d)} = \frac{1}{17}$$

Cross-multiplying gives us $17(a + 2d) = 9(a + 4d)$, which simplifies to $17a + 34d = 9a + 36d$. Rearranging the terms, we find $8a = 2d$, or simply $d = 4a$.

This is the hidden geometric reality of our sequence: the common difference is exactly four times the first term!

### Phase 3: The Lock and Key

Now, we use the constraint $110 < a_{15} < 120$. The general term formula is $a_n = a + (n-1)d$.

For the 15th term, $a_{15} = a + 14d$. Substituting our relationship $d = 4a$, we get:

$$a_{15} = a + 14(4a) = a + 56a = 57a$$

Our inequality becomes $110 < 57a < 120$. This is where the 'natural number' constraint shines. We test values for $a$:

- If $a=1$, $57(1) = 57$ (too small).
- If $a=2$, $57(2) = 114$ (perfect! $110 < 114 < 120$).
- If $a=3$, $57(3) = 171$ (too large).

Thus, $a$ must be $2$, and consequently, $d = 4(2) = 8$.

### Phase 4: The Final Calculation

We have arrived at the finish line. We need the sum of the first ten terms, $S_{10}$:

$$S_{10} = \frac{10}{2} [2a + (10-1)d] = 5 [2(2) + 9(8)]$$

$$S_{10} = 5 [4 + 72] = 5 [76] = 380$$

And there it is! The **sum is 380**. By respecting the constraints and simplifying the algebra, we turned a complex-looking problem into a clear, logical path.

Keep practicing this mindset—the math is always simpler than it appears if you look for the structure!

Select Answer:

Visualized Solution

Define the A.P. Parameters

Recall the Sum Formula $S_{n}$

Set up the Ratio $\frac{S _{5}}{S _{9}}$

Substitute into the Ratio

Simplify the Ratio Equation

Solve for $d$ in terms of $a$

Find the 15th Term $a_{15}$

Apply the Constraint $110 < a_{15} < 120$

Determine $a$ and $d$

Calculate $S_{10}$

Final Computation

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

The Beauty of Arithmetic Progressions

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Let $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100$ . For any integer $n$ with $1 \leq n \leq 20$ , let $m = 5 n$ . If $S_{m} / S_{n}$ does not depend on $n$ , then $a_{2}$ is \dots.

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Select Answer:

Visualized Solution

Define the A.P. Parameters

Recall the Sum Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Set up the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S9​S5​​