If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the la

Q: Why is the total number of terms $n+2$ and not $n$?

Great question! When you insert $n$ terms between two existing numbers, you are adding $n$ new terms to the original two. So, the total count becomes $n$ (the means) + $2$ (the original endpoints) = $n+2$ terms.

Q: Why can't we just use the sum formula for arithmetic means?

The sum formula is useful for finding the total sum of the sequence, but here we are dealing with the ratio of specific terms. Focusing on the general term formula $T_m = a + (m-1)d$ is much more direct and avoids unnecessary complexity.

Q: How do I know which root of the quadratic equation to pick?

In physics and mathematics, we often get multiple roots. However, $n$ represents the number of terms, which must be a positive integer. If a root is negative or a fraction, it physically cannot represent a count of terms, so we discard it.

Question

If $n$ arithmetic means are inserted between $a$ and 100 such that the ratio of the first mean to the last mean is $1 : 7$ and $a + n = 33$, then the value of $n$ is

Options:

(A) 21

(B) 22

(C) 23

(D) 24

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a number line, looking at two points: $a$ and $100$. Between them, we are tasked with placing $n$ arithmetic means. This creates a sequence that flows with a constant rhythm: $a, A_1, A_2, \dots, A_n, 100$.

The total number of terms in this sequence is $n+2$, as we have $n$ means plus the two endpoints. This count serves as the foundation of our entire structure.

### The Heartbeat of the Progression

Every arithmetic progression has a heartbeat: the common difference, $d$. We know the formula for the $m$-th term of an AP is $T_m = a + (m-1)d$.

Since $100$ is the $(n+2)$-th term, we can write:

$$100 = a + (n+2-1)d$$

This simplifies to $100 = a + (n+1)d$. By isolating $d$, we find the step size required to move from one mean to the next:

$$d = \frac{100-a}{n+1}$$

### The Ratio: A Window into the Sequence

We are given a fascinating clue: the ratio of the first mean $A_1$ to the last mean $A_n$ is $1:7$. We express these means in terms of $a$ and $d$:

$$A_1 = a + d = a + \frac{100-a}{n+1} = \frac{an+100}{n+1}$$

$$A_n = 100 - d = 100 - \frac{100-a}{n+1} = \frac{100n+a}{n+1}$$

When we set up the ratio $\frac{A_1}{A_n} = \frac{1}{7}$, the denominators $(n+1)$ cancel out, leaving us with:

$$\frac{an+100}{100n+a} = \frac{1}{7}$$

### The Final Calculation

Cross-multiplying the ratio gives us $7(an+100) = 100n+a$, which expands to $7an + 700 = 100n + a$. We are given the constraint $a+n=33$, or $a = 33-n$.

Substituting $a = 33-n$ into our equation:

$$7(33-n)n + 700 = 100n + (33-n)$$

Expanding this expression yields:

$$231n - 7n^2 + 700 = 99n + 33$$

Rearranging all terms to one side, we form the quadratic equation:

$$7n^2 - 132n - 667 = 0$$

Solving this quadratic equation, we find the roots. Since $n$ must be a positive integer, we discard the negative result and conclude that **$n=23$**.

Animated Solution for Mathematics - Sequence and Series: If $n$ arithmetic means are inserted between $a$ and 100 such that the ratio of the first mean to the last mean is $1 : 7$ and $a + n = 33$ , then the value of $n$ is

Select Answer:

Visualized Solution

Visualizing the Sequence

The Common Difference $d$

Expressing $A_{1}$ and $A_{n}$

Simplifying $A_{1}$ and $A_{n}$

Applying the Given Ratio

Cross-Multiplication

Substitution of $a$

Expanding the Expression

Forming the Quadratic Equation

Solving for $n$

Final Conclusion

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

Analyzing the Setup

The Heartbeat of the Progression

The Ratio

The Final Calculation

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