Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference

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Concept understanding — Mean Variance Natural Numbers

Mean and Variance of Natural Numbers

Let’s start with something you already know: the mean (average) and variance (spread) of a set of numbers. If I give you the first five natural numbers — 1, 2, 3, 4, 5 — you can compute their mean and variance easily. But what if I ask: What is the mean of all natural numbers? That’s infinite, so it doesn’t make sense directly. Instead, we ask: What is the mean of the first $n$ natural numbers? And then we see how it behaves as $n$ grows.

That’s the core idea: we study the mean and variance of the first $n$ natural numbers as a function of $n$ , and often look at what happens when $n$ becomes very large.

Intuition First

Imagine you line up the numbers $1, 2, 3, \dots, n$ on a number line. Their average is somewhere in the middle — roughly $n/2$ . More precisely, the mean of the first $n$ natural numbers is $\frac{n+1}{2}$ . For $n=5$ , that’s $3$ , which matches your intuition.

Now, variance measures how spread out the numbers are around that mean. For small $n$ , the spread is small; for large $n$ , the spread grows. The variance of the first $n$ natural numbers turns out to be $\frac{n^2 - 1}{12}$ . For $n=5$ , that’s $\frac{25-1}{12} = 2$ , which is a moderate spread.

Note

These formulas assume we are using population variance (dividing by $n$ , not $n-1$ ). In exam contexts, always check which variance definition is expected — but for natural numbers, population variance is standard.

Precise Statement

Let $X$ be a random variable that takes values $1, 2, 3, \dots, n$ with equal probability $1/n$ . Then:

\text{Mean: } \mu_n = \frac{n+1}{2}

These are exact formulas for any positive integer $n$ .

Derivation (Why These Formulas?)

Mean

The sum of the first $n$ natural numbers is $1 + 2 + \dots + n = \frac{n(n+1)}{2}$ .

Since there are $n$ numbers, the mean is:

\mu_n = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}

Variance

Variance is the average of squared deviations from the mean:

\sigma_n^2 = \frac{1}{n} \sum_{k=1}^n (k - \mu_n)^2

A cleaner way uses the identity: $\sigma^2 = E[X^2] - (E[X])^2$ .

First, $E[X^2] = \frac{1}{n} \sum_{k=1}^n k^2$ . The sum of squares formula is $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ . So:

E[X^2] = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}

Now, $(E[X])^2 = \left( \frac{n+1}{2} \right)^2 = \frac{(n+1)^2}{4}$ .

Therefore:

\sigma_n^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}

Factor $(n+1)$ :

\sigma_n^2 = (n+1) \left[ \frac{2n+1}{6} - \frac{n+1}{4} \right]

Compute the bracket: common denominator 12:

\frac{2(2n+1) - 3(n+1)}{12} = \frac{4n+2 - 3n - 3}{12} = \frac{n - 1}{12}

Thus:

\sigma_n^2 = (n+1) \cdot \frac{n-1}{12} = \frac{n^2 - 1}{12}

What This Tells You

The mean grows linearly with $n$ — roughly half of $n$ .
The variance grows quadratically — roughly $n^2/12$ for large $n$ .
For large $n$ , the standard deviation $\sigma_n \approx \frac{n}{\sqrt{12}} \approx 0.2887n$ , meaning the spread is about 29% of the range. …

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The mean of the first $n$ terms of an AP is the average of the first and last terms, $\frac{2a+(n-1)d}{2}$ , and the standard deviation is $\frac{|d|}{\sqrt{12}}\sqrt{n^2-1}$ — a result that depends only on the common difference and $n$ , not on $a$ .

The key insight here is that an arithmetic progression is just a set of equally spaced numbers. When you shift all numbers by a constant (the first term $a$ ), the mean shifts by that constant, but the standard deviation — which measures spread — stays unchanged. The spread depends only on the spacing $d$ and how many terms $n$ you take.

Let’s build this from the ground up.

1. Write down the terms

The first $n$ terms of an AP with first term $a$ and common difference $d$ are:

a,\ a+d,\ a+2d,\ \dots,\ a+(n-1)d

2. Find the mean (arithmetic mean)

The mean $\bar{x}$ is the sum divided by $n$ .

Sum of an AP:

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

So the mean is:

\bar{x} = \frac{S_n}{n} = \frac{2a + (n-1)d}{2}

Tip

Notice this is exactly the average of the first term $a$ and the last term $a+(n-1)d$ . That’s always true for any AP — the mean of equally spaced numbers is the midpoint of the extremes.

3. Shift the data to simplify variance

Standard deviation is unchanged if we subtract a constant from every term. Subtract the mean $\bar{x}$ from each term. This gives a new set of numbers centered at zero:

Let $t_k = a + (k-1)d - \bar{x}$ for $k = 1, 2, \dots, n$ .

Since $\bar{x} = a + \frac{(n-1)d}{2}$ , we have:

t_k = a + (k-1)d - a - \frac{(n-1)d}{2} = \left(k-1 - \frac{n-1}{2}\right)d

So:

t_k = d\left(k - \frac{n+1}{2}\right)

These are symmetric about zero: the terms are $-\frac{n-1}{2}d,\ -\frac{n-3}{2}d,\ \dots,\ 0,\ \dots,\ \frac{n-3}{2}d,\ \frac{n-1}{2}d$ (if $n$ is odd, the middle term is exactly 0; if $n$ is even, the two middle terms are $\pm\frac{d}{2}$ ).

4. Variance = mean of squared deviations

Variance $\sigma^2$ is:

\sigma^2 = \frac{1}{n}\sum_{k=1}^n t_k^2 = \frac{d^2}{n}\sum_{k=1}^n \left(k - \frac{n+1}{2}\right)^2

Let $m = \frac{n+1}{2}$ . Then we need:

\sum_{k=1}^n (k - m)^2

This is a standard sum. Expand:

\sum_{k=1}^n (k^2 - 2mk + m^2) = \sum k^2 - 2m\sum k + n m^2

We know:

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$
$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Substitute $m = \frac{n+1}{2}$ :

\sum k^2 = \frac{n(n+1)(2n+1)}{6}

2m\sum k = 2\cdot\frac{n+1}{2}\cdot\frac{n(n+1)}{2} = \frac{n(n+1)^2}{2}

n m^2 = n\cdot\frac{(n+1)^2}{4} = \frac{n(n+1)^2}{4}

So:

\sum (k-m)^2 = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)^2}{2} + \frac{n(n+1)^2}{4}

Combine the last two terms: $-\frac{1}{2} + \frac{1}{4} = -\frac{1}{4}$ , so:

= \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)^2}{4}

Factor $\frac{n(n+1)}{12}$ : …

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Q.Determine mean and standard deviation of first nnn terms of an A.P. whose first term is aaa and common difference is ddd.

1. Write down the terms

2. Find the mean (arithmetic mean)

3. Shift the data to simplify variance

4. Variance = mean of squared deviations

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Q.Determine mean and standard deviation of first $n$ terms of an A.P. whose first term is $a$ and common difference is $d$ .

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