Arithmetic Sequences

From repeated addition to the sum of any sequence — discover the pattern yourself

A Repeated addition — the simplest sequence

The simplest sequence of all: the same number repeated over and over. Before tackling harder sums, make sure you can handle this warm-up confidently.

A1 Find the value of $\underbrace{6 + 6 + 6 + \cdots + 6}_{10 \text{ times}}$ Starter ▼

You add the number $6$ exactly $10$ times: $$6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 = \;?$$

Hint: Count the terms. Adding the same number many times is the definition of multiplication. How does that help?

Show solution

There are 10 copies of 6.

$10 \times 6 = 60$

$= 60$

A2 Find $\underbrace{7 + 7 + 7 + \cdots + 7}_{6 \text{ times}}$ Starter ▼

$$7 + 7 + 7 + 7 + 7 + 7 = \;?$$

Hint: How many 7s are there?

Show solution

$6 \times 7 = 42$

$= 42$

A3 If you add the number $k$ exactly $n$ times, what is the sum? Easy ▼

Write a single formula for: $$\underbrace{k + k + k + \cdots + k}_{n \text{ times}} = \;?$$

Hint: What pattern did you see in A1 and A2? Replace the specific numbers with $k$ and $n$.

Show solution

$$\underbrace{k + k + \cdots + k}_{n} = n \times k$$

This works because multiplication is repeated addition. We will need this fact in Section C.

B Adding consecutive numbers — finding a shortcut

Now the numbers are no longer all the same — they grow by 1 each step. Start small, where direct addition is easy. Then see if you can spot a pattern that makes the large cases manageable without adding every term.

B1 Find $1 + 2 + 3 + 4 + 5$ Starter ▼

Add the first five positive integers: $$1 + 2 + 3 + 4 + 5 = \;?$$

Hint: These are small enough to add directly. While you compute, notice: what is $1 + 5$? What is $2 + 4$? Do you see anything?

Show solution

$1 + 2 = 3$

$3 + 3 = 6$

$6 + 4 = 10$

$10 + 5 = 15$

$= 15$

Notice: $1 + 5 = 6$, $\;2 + 4 = 6$, $\;3 = 3$ (the middle). Three groups of 6 minus one 3? Or five terms averaging $3$? Keep this in mind.

B2 Find $1 + 2 + 3 + \cdots + 10$ Easy ▼

Add all integers from 1 to 10: $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = \;?$$

Hint: You could add these one by one. But first ask: what is $1 + 10$? What is $2 + 9$? What is $3 + 8$? How many such pairs are there, and what does each pair sum to?

Show solution

Pair up first and last: $1+10 = 11$, $\ 2+9 = 11$, $\ 3+8 = 11$, $\ 4+7 = 11$, $\ 5+6 = 11$

There are $5$ pairs, each summing to $11$.

$5 \times 11 = 55$

$= 55$

The pairing trick — why it works

Here is a slicker way to see the same idea. Write the sum $S$ twice: once forwards, once backwards. Stack them and add column by column.

S = 1 + 2 + 3 + 4 + 5 + 6
S = 6 + 5 + 4 + 3 + 2 + 1
── ── ── ── ── ── ── 2S = 7 + 7 + 7 + 7 + 7 + 7
2S = 6 × 7 = 42 ⟹ S = 21

Each column sums to the same value: $(\text{first} + \text{last}) = 1 + 6 = 7$. There are $6$ columns (one per number). So $2S = 6 \times 7$, giving $S = 21$.

You don't need to list all the pairs. You just need to know how many numbers there are and what the first and last are.

B3 Find $1 + 2 + 3 + \cdots + 20$ Easy ▼

Add all integers from 1 to 20: $$1 + 2 + 3 + \cdots + 20 = \;?$$

S = 1 + 2 + 3 + ··· + 18 + 19 + 20
S = 20 + 19 + 18 + ··· + 3 + 2 + 1
── ── ── ── ── ── ── 2S = 21 + 21 + 21 + ··· + 21 + 21 + 21

Hint: How many columns are in the addition above? What does each column sum to? Once you know $2S$, divide by 2.

Show solution

There are $20$ columns (one for each number from 1 to 20).

Each column sums to $1 + 20 = 21$.

$2S = 20 \times 21 = 420$

$S = \dfrac{420}{2} = 210$

$1 + 2 + \cdots + 20 = 210$

B4 Find $1 + 2 + 3 + \cdots + 50$ Medium ▼

Add all integers from 1 to 50. No calculator needed — use the trick.

Hint: What is $1 + 50$? How many columns will you have? Set up $2S = (\text{number of terms}) \times (\text{column sum})$.

Show solution

Number of terms: $50$.

Each column: $1 + 50 = 51$.

$2S = 50 \times 51 = 2550$

$S = \dfrac{2550}{2} = 1275$

$1 + 2 + \cdots + 50 = 1275$

B5 Find $1 + 2 + 3 + \cdots + 100$ Medium ▼

Add all integers from 1 to 100. Legend has it a young student named Gauss solved this in seconds as a school task. Can you match him?

Hint: Same method. How many terms? What does each paired column sum to?

Show solution

Number of terms: $100$.

Each column: $1 + 100 = 101$.

$2S = 100 \times 101 = 10100$

$S = \dfrac{10100}{2} = 5050$

$1 + 2 + \cdots + 100 = 5050$

This is the answer Gauss supposedly gave instantly as a child. The trick is elegant: no matter how large $n$ is, the computation stays just as simple.

B6 Find $1 + 2 + 3 + \cdots + 1000$ Medium ▼

Add all integers from 1 to 1000. A direct sum would take a very long time — but with the trick, it takes three steps.

Hint: $2S = 1000 \times (1 + 1000)$.

Show solution

Number of terms: $1000$.

Each column: $1 + 1000 = 1001$.

$2S = 1000 \times 1001 = 1\,001\,000$

$S = \dfrac{1\,001\,000}{2} = 500\,500$

$1 + 2 + \cdots + 1000 = 500\,500$

C Discovering the general formula

You applied the pairing trick to specific values of $n$. Now do the same thing with $n$ left as a letter — and you will derive a formula that works for any $n$ in one step.

C1 Derive a formula for $1 + 2 + 3 + \cdots + n$ Medium ▼

Let $S = 1 + 2 + 3 + \cdots + n$ where $n$ is any positive integer. Apply the forwards-and-backwards trick to derive a closed formula for $S$.

S = 1 + 2 + 3 + ··· + (n−2) + (n−1) + n
S = n + (n−1) + (n−2) + ··· + 3 + 2 + 1
── ────── ────── ────── ────── ── 2S = (n+1) + (n+1) + (n+1) + ··· + (n+1) + (n+1) + (n+1)

Hint: How many columns are in the table above? What does each column sum to? Write an expression for $2S$, then solve for $S$.

Show derivation

There are $n$ columns (one per term).

Each column sums to $1 + n = n + 1$.

$2S = n \times (n+1)$

$S = \dfrac{n(n+1)}{2}$

$$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$

You derived this — not memorised it. The formula works because: $n$ terms, average value $\dfrac{n+1}{2}$, total $= n \times \dfrac{n+1}{2}$.

C2 Verify: check the formula against B1 – B6. Easy ▼

Use the formula $S = \dfrac{n(n+1)}{2}$ to recompute the answers from exercises B1 through B6. Do they match?

Hint: Substitute $n = 5, 10, 20, 50, 100, 1000$ in turn.

Show verification

$n = 5:\quad \dfrac{5 \times 6}{2} = 15$ ✓

$n = 10:\quad \dfrac{10 \times 11}{2} = 55$ ✓

$n = 20:\quad \dfrac{20 \times 21}{2} = 210$ ✓

$n = 50:\quad \dfrac{50 \times 51}{2} = 1275$ ✓

$n = 100:\quad \dfrac{100 \times 101}{2} = 5050$ ✓

$n = 1000:\quad \dfrac{1000 \times 1001}{2} = 500\,500$ ✓

All match.

C3 Find $1 + 2 + \cdots + 500$ Easy ▼

Use your formula to find $1 + 2 + 3 + \cdots + 500$. You should be able to write the answer in one line.

Show solution

$S = \dfrac{500 \times 501}{2} = 250 \times 501 = 125\,250$

$= 125\,250$

C4 Find $3 + 7 + 11 + 15 + 19 + 23$ Medium ▼

Add up this sequence: $3 + 7 + 11 + 15 + 19 + 23$. Each term is 4 more than the previous one. Can you use the same pairing trick?

Hint: Try pairing first and last terms: $3 + 23 = ?$ Then second and second-to-last: $7 + 19 = ?$ Then $11 + 15 = ?$ How many pairs are there?

Show solution

$3 + 23 = 26$

$7 + 19 = 26$

$11 + 15 = 26$

Three pairs, each summing to $26$: $\;3 \times 26 = 78$

$3 + 7 + 11 + 15 + 19 + 23 = 78$

The pairing trick works even when the sequence doesn't start at 1! The key observation: every pair summed to the same value — the first term plus the last term.

C5 Derive a formula for any arithmetic sequence. Hard ▼

An arithmetic sequence starts at value $a$, ends at value $\ell$, and has $n$ terms in total (each step increases by the same amount $d$): $$S = a + (a+d) + (a+2d) + \cdots + (\ell - d) + \ell$$ Apply the forwards-and-backwards trick to find a formula for $S$ in terms of $a$, $\ell$, and $n$.

S = a + (a+d) + (a+2d) + ··· + (ℓ−d) + ℓ
S = ℓ + (ℓ−d) + (ℓ−2d) + ··· + (a+d) + a
── ────── ────── ────── ── 2S = (a+ℓ) + (a+ℓ) + (a+ℓ) + ··· + (a+ℓ) + (a+ℓ)

Hint: What does each column sum to? How many columns are there? Write $2S$ as a product, then divide by 2.

Show derivation

Each column sums to $a + \ell$.

There are $n$ columns (one per term).

$2S = n \times (a + \ell)$

$$S = \frac{n}{2}(a + \ell) = \frac{n}{2}(\text{first} + \text{last})$$

This is the most general form of the sum formula for an arithmetic sequence. When $a = 1$ and $\ell = n$, it reduces to $\dfrac{n}{2}(1 + n) = \dfrac{n(n+1)}{2}$, which is exactly what you derived in C1.

C6 Find $5 + 10 + 15 + \cdots + 100$ Hard ▼

Find the sum of multiples of 5 from 5 to 100: $$5 + 10 + 15 + \cdots + 100$$ First work out how many terms there are, then apply the formula from C5.

Hint: The terms are $5 \times 1, \; 5 \times 2, \; \ldots, \; 5 \times 20$. How many terms? What are $a$ and $\ell$?

Show solution

Terms: $5, 10, 15, \ldots, 100$ — that is $20$ terms ($n = 20$).

First term: $a = 5$. Last term: $\ell = 100$.

$S = \dfrac{20}{2}(5 + 100) = 10 \times 105 = 1050$

$5 + 10 + \cdots + 100 = 1050$

Alternatively: $5 + 10 + \cdots + 100 = 5(1 + 2 + \cdots + 20) = 5 \times 210 = 1050$ ✓

C7 Rewrite the formula using first term $a$, step size $d$, and count $n$. Hard ▼

In C5 you found $S = \dfrac{n}{2}(a + \ell)$. But sometimes you don't know the last term $\ell$ — you only know the first term $a$, the common difference $d$, and the number of terms $n$. Express $\ell$ in terms of $a$, $d$, and $n$, and substitute into the formula.

Hint: The sequence is $a,\; a+d,\; a+2d,\; \ldots$ The $k$-th term (starting from $k=1$) is $a + (k-1)d$. What is the $n$-th term?

Show derivation

$n$-th term: $\ell = a + (n-1)d$

Substitute into $S = \dfrac{n}{2}(a + \ell)$: $$S = \frac{n}{2}\bigl(a + a + (n-1)d\bigr) = \frac{n}{2}\bigl(2a + (n-1)d\bigr)$$

$$S = \frac{n}{2}\bigl(2a + (n-1)d\bigr)$$

Quick check: For $a=1$, $d=1$, $n$ terms: $S = \dfrac{n}{2}(2 + n - 1) = \dfrac{n(n+1)}{2}$ ✓

C8 Find the sum of the first 15 terms of $4, 7, 10, 13, \ldots$ Hard ▼

The sequence $4, 7, 10, 13, \ldots$ has first term $a = 4$ and common difference $d = 3$. Find the sum of the first 15 terms.

Hint: Use $S = \dfrac{n}{2}\bigl(2a + (n-1)d\bigr)$ with $n=15$, $a=4$, $d=3$.

Show solution

$n = 15,\quad a = 4,\quad d = 3$

$2a + (n-1)d = 8 + 14 \times 3 = 8 + 42 = 50$

$S = \dfrac{15}{2} \times 50 = 15 \times 25 = 375$

Sum of first 15 terms $= 375$

Check: last term $= 4 + 14 \times 3 = 46$. Using the other formula: $S = \dfrac{15}{2}(4 + 46) = \dfrac{15 \times 50}{2} = 375$ ✓