Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P.

Worked Examples · Example 14

Q. Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P.

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Concept understanding — Arithmetic Mean

What is the Arithmetic Mean?

Imagine you and three friends score the following marks in a test: 6, 8, 10, and 12. If you had to pick a single number that "represents" the group's performance, what would it be? The natural idea is to share the total equally among everyone. The total marks are $6 + 8 + 10 + 12 = 36$ . If you split this equally among four people, each gets $36 \div 4 = 9$ . That 9 is the arithmetic mean — the "fair share" of the total.

The arithmetic mean answers the question: If everyone had the same value, what would that value be? It is the number that, when repeated for each observation, gives the same total as the original data.

The Precise Definition

For a set of $n$ numbers $x_1, x_2, x_3, \dots, x_n$ , the arithmetic mean (often just called the mean or average) is:

\bar{x} = \frac{x_1 + x_2 + x_3 + \dots + x_n}{n}

The symbol $\bar{x}$ (read "x-bar") is standard notation for the mean of a sample. The numerator is the sum of all observations, and the denominator is the count of observations.

Why It Works

The mean has a beautiful property: the total deviation from the mean is always zero. Take the example above: the deviations are $6-9 = -3$ , $8-9 = -1$ , $10-9 = +1$ , $12-9 = +3$ . Their sum is $(-3) + (-1) + 1 + 3 = 0$ . This is not a coincidence — it is built into the definition. The mean is the unique point where the positive and negative differences cancel out perfectly.

Tip

This "balancing point" property is why the mean is sensitive to extreme values. If one student scored 100 instead of 12, the mean would jump dramatically — the "fair share" gets pulled toward the outlier.

A Quick Example

Find the arithmetic mean of $15, 22, 18, 30$ .

Step 1: Sum the numbers: $15 + 22 + 18 + 30 = 85$

Step 2: Count the numbers: $n = 4$

Step 3: Divide: $\frac{85}{4} = 21.25$

Important

Answer: $\bar{x} = 21.25$

Common Pitfall

Do not confuse the arithmetic mean with the median (the middle value when sorted) or the mode (the most frequent value). The mean uses every data point in its calculation, which is both its strength (it uses all information) and its weakness (it can be distorted by outliers). For the set $1, 2, 3, 100$ , the mean is $26.5$ , but the median is $2.5$ — a much better "typical" value here.

The arithmetic mean is the most widely used measure of central tendency, but always ask: Does "fair share" make sense for this data?

Inserting 6 numbers between 3 and 24 creates an 8-term A.P.; solving for the common difference gives $d=3$ , so the inserted numbers are 6, 9, 12, 15, 18, 21.

k

We insert $k=6$ numbers between $a=3$ and $b=24$ , so the total number of terms (including both endpoints) is $n = 6+2 = 8$ .
Apply the formula: $24 = 3+(8-1)d = 3+7d$ .
Solve for $d$ : $7d = 24-3 = 21 \Rightarrow d = 3$ .
Generate the 8-term A.P. starting at $3$ with common difference $3$ : $3,\,6,\,9,\,12,\,15,\,18,\,21,\,24$ .
The first and last terms ( $3$ and $24$ ) are the given endpoints; the 6 terms in between are the inserted numbers.
Self-check: the sequence has exactly 8 terms as required, common difference is constant ( $6-3=3$ , $9-6=3$ , …, $24-21=3$ ), and it starts at 3 and ends at 24. ✓

✓Final answer

\mathbf{6, 9, 12, 15, 18, 21}

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Q.Insert 6 numbers between 3 and 24 such that the resulting sequence is an A.P.

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