Sum of the first $n$ terms of an AP

A famous schoolboy story claims that when young Gauss was asked to add $1 + 2 + 3 + \ldots + 100$, he wrote $5050$ within seconds. His trick , pair the first and last terms, then the second and second-last, and so on , gives the formula we need.

The formula

For an AP with first term $a$, common difference $d$, and $n$ terms, the sum of the first $n$ terms is $\boxed{S_n = \frac{n}{2} \bigl[2 a + (n - 1) d\bigr].}$

Equivalently, if the last term is $\ell = a + (n - 1) d$, $S_n = \frac{n}{2} (a + \ell).$

Derivation (Gauss trick)

Write the sum forwards and backwards: $S_n = a + (a + d) + (a + 2 d) + \ldots + \ell,$ $S_n = \ell + (\ell - d) + (\ell - 2 d) + \ldots + a.$ Add term by term: $2 S_n = (a + \ell) + (a + \ell) + \ldots + (a + \ell) = n (a + \ell).$ So $S_n = \dfrac{n}{2} (a + \ell)$.

Substituting $\ell = a + (n - 1) d$ gives the equivalent formula $S_n = \frac{n}{2} \bigl[2 a + (n - 1) d\bigr].$

When to use which form

• $S_n = \dfrac{n}{2}(a + \ell)$: ideal when $a$ and $\ell$ are given (and so we just need $n$).
• $S_n = \dfrac{n}{2}[2 a + (n - 1) d]$: ideal when $a$ and $d$ are given (so we don't have to compute the last term first).

Both are equivalent , choose whichever leads to less arithmetic.

A useful corollary: the $n$th term and the sum are linked by $a_n = S_n - S_{n - 1}.$ This is occasionally handy: if you are given the sum formula and need the $n$th term, just subtract.

Worked examples

Example 1. Sum of the first $30$ terms of $5, 9, 13, 17, \ldots$.

$a = 5, d = 4, n = 30$. $S_{30} = \dfrac{30}{2}[2 \cdot 5 + 29 \cdot 4] = 15 \cdot [10 + 116] = 15 \cdot 126 = 1890$.

Example 2. Find the sum of the first $50$ natural numbers.

$1 + 2 + \ldots + 50 = \dfrac{50}{2}(1 + 50) = 25 \cdot 51 = 1275$.

Example 3. Find the sum of all two-digit multiples of $3$.

Two-digit multiples of $3$: $12, 15, \ldots, 99$. AP with $a = 12, d = 3$.

Number of terms: $a + (n - 1) d = 99 \Rightarrow 12 + 3(n - 1) = 99 \Rightarrow n = 30$.

Sum $= \dfrac{30}{2}(12 + 99) = 15 \cdot 111 = 1665$.

Example 4. How many terms of the AP $9, 17, 25, \ldots$ are required to give a sum of $636$?

$a = 9, d = 8$. $S_n = \dfrac{n}{2}[18 + 8(n - 1)] = \dfrac{n}{2}(8 n + 10) = n(4 n + 5)$.

Set $4 n^2 + 5 n = 636 \Rightarrow 4 n^2 + 5 n - 636 = 0$. Discriminant $= 25 + 10176 = 10201 = 101^2$. $n = (-5 + 101)/8 = 12$.

Example 5. The sum of the first $n$ terms of an AP is $4 n^2 + 5 n$. Find the $n$th term.

$a_n = S_n - S_{n-1} = (4 n^2 + 5 n) - (4 (n - 1)^2 + 5 (n - 1)) = (4 n^2 + 5 n) - (4 n^2 - 8 n + 4 + 5 n - 5) = (4 n^2 + 5 n) - (4 n^2 - 3 n - 1) = 8 n + 1$.

So $a_n = 8 n + 1$, an AP with $a = 9, d = 8$. (And indeed $S_1 = a_1 = 9$; check $S_1 = 4 + 5 = 9$ ✓.)

Try it yourself

1. Find $S_{15}$ for $1, 4, 7, 10, \ldots$.
2. Find the sum of first $100$ positive integers.
3. Sum of all odd numbers from $1$ to $99$.
4. Sum of all multiples of $7$ between $1$ and $100$.
5. The sum of the first $n$ terms of an AP is $3 n^2 + 2 n$. Find the AP.
6. If $S_n$ for an AP is $\dfrac{n}{2}(3 n + 5)$, find the AP.
7. How many terms of the AP $24, 21, 18, \ldots$ must be taken so that the sum is $78$?
8. Find the sum of all three-digit numbers divisible by $9$.
9. The first term of an AP is $5$, the last term is $45$, and the sum is $400$. Find the number of terms and the common difference.
10. A man arranges to pay a debt of ₹$3600$ by $40$ instalments which form an AP. When $30$ instalments are paid he dies, leaving one-third of the debt unpaid. Find the value of the first instalment.

Pitfalls / Insight

• Don't forget the $n/2$ in front. The most common error is writing $S_n = 2 a + (n - 1) d$ (missing the multiplier).
• For "how many terms", the quadratic in $n$ should give a positive integer , reject negatives or fractions.
• For "sum of all multiples" in a range, first identify the AP, then count terms, then sum.

Insight. Once you have $a$ and $d$, both the $n$th term and the sum are one-line computations. Spend your effort getting $a$ and $d$ from the problem; the rest is arithmetic.