Sum of First n Terms of an AP

The Sum Formula

Adding many terms of an AP one by one is slow. There's a beautiful shortcut. The sum of the first $n$ terms of an AP, written $S_n$, is:

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

where $a$ is the first term, $d$ the common difference, and $n$ the number of terms.

The story behind it

Legend says the young mathematician Gauss added $1 + 2 + \dots + 100$ instantly by pairing the first and last terms. Each pair sums to the same value, and this idea gives the formula.

Key Point: $S_n = \dfrac{n}{2}[2a + (n-1)d]$. You need $a$, $d$, and $n$.

[Board Important] Don't confuse $S_n$ (sum of $n$ terms) with $a_n$ (the $n$th term). $S_n$ adds up all terms up to the $n$th.

The Alternative Sum Formula

When the last term $l$ is known, there's a simpler version:

$$S_n = \frac{n}{2}(a + l)$$

This says: the sum equals the number of terms times the average of the first and last term.

Why it's the same

Since $l = a_n = a + (n-1)d$, substituting into $\dfrac{n}{2}(a + l)$ recovers the first formula.

Worked outline

Find $1 + 2 + 3 + \dots + 100$: here $a = 1$, $l = 100$, $n = 100$. $$S_{100} = \frac{100}{2}(1 + 100) = 50 \times 101 = 5050.$$

Key Point: Use $S_n = \dfrac{n}{2}(a + l)$ when the last term is given; use $S_n = \dfrac{n}{2}[2a + (n-1)d]$ when $d$ is given.

[Board Important] The 'first + last' average formula is the fastest when you know the last term — a real time-saver.

Finding a Term from the Sum

There's a neat link between sums and terms:

$$a_n = S_n - S_{n-1}$$

That is, the $n$th term equals the sum of the first $n$ terms minus the sum of the first $(n-1)$ terms.

When $S_n$ is given as a formula

If a problem gives $S_n = 3n^2 + 5n$, then:

• $S_1 = a_1 = 8$ (first term).
• $a_n = S_n - S_{n-1}$ gives the general term, and $d$ follows.

Key Point: $a_n = S_n - S_{n-1}$ lets you recover terms (and $a$, $d$) directly from a sum formula.

[Board Important] If $S_n$ is a quadratic in $n$ (like $3n^2 + 5n$), the list is automatically an AP, and the coefficient pattern tells you $a$ and $d$.

Finding the Number of Terms from a Sum

Often a problem gives the sum and asks how many terms produce it. Set $S_n$ equal to the given sum — this gives a quadratic in $n$ to solve.

Worked outline

How many terms of $24, 21, 18, \dots$ give a sum of 78? Here $a = 24$, $d = -3$. $$\frac{n}{2}[48 + (n-1)(-3)] = 78 \Rightarrow n(51 - 3n) = 156 \Rightarrow 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0.$$ Factoring: $(n - 4)(n - 13) = 0$, so $n = 4$ or $n = 13$. Both can be valid (the AP turns negative and the later terms cancel out).

Key Point: Setting $S_n$ equal to a value gives a quadratic in $n$. Sometimes two values of $n$ both work — interpret carefully.

[Board Important] Reject any non-positive or non-integer $n$. If both roots are valid positive integers, mention both.

Solved Examples

Example 1: Sum using $a$, $d$, $n$

Find the sum of the first 22 terms of $8, 3, -2, \dots$

Solution:

1. $a = 8$, $d = -5$, $n = 22$.
2. $S_{22} = \dfrac{22}{2}[16 + 21(-5)] = 11[16 - 105] = 11(-89) = -979$.

Final Answer: $-979$.

Takeaway: Apply $S_n = \dfrac{n}{2}[2a + (n-1)d]$ carefully with signs.

Example 2: Sum using first and last

Find $1 + 2 + 3 + \dots + 100$.

Solution:

1. $a = 1$, $l = 100$, $n = 100$.
2. $S_{100} = \dfrac{100}{2}(1 + 100) = 50 \times 101 = 5050$.

Final Answer: 5050.

Takeaway: $S_n = \dfrac{n}{2}(a + l)$ is fastest with a known last term.

Example 3: Sum of first 1000 positive integers

Find the sum of the first 1000 positive integers.

Solution:

1. $a = 1$, $l = 1000$, $n = 1000$.
2. $S = \dfrac{1000}{2}(1 + 1000) = 500 \times 1001 = 500500$.

Final Answer: 500500.

Takeaway: Same average-based formula scales to any range.

Example 4: How many terms give a sum?

How many terms of $24, 21, 18, \dots$ are needed to give a sum of 78?

Solution:

1. $a = 24$, $d = -3$. $\dfrac{n}{2}[48 + (n-1)(-3)] = 78$.
2. $n(51 - 3n) = 156 \Rightarrow 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0$.
3. $(n - 4)(n - 13) = 0 \Rightarrow n = 4$ or $13$.

Final Answer: 4 or 13 terms.

Takeaway: Setting $S_n = 78$ gives a quadratic; both positive roots are valid here.

Example 5: Sum of first $n$ natural numbers

Find the sum of the first $n$ natural numbers in terms of $n$.

Solution:

1. $a = 1$, $l = n$, number of terms $= n$.
2. $S_n = \dfrac{n}{2}(1 + n) = \dfrac{n(n+1)}{2}$.

Final Answer: $\dfrac{n(n+1)}{2}$.

Takeaway: The famous formula $1 + 2 + \dots + n = \dfrac{n(n+1)}{2}$.

Example 6: Find $a_n$ from $S_n$

If $S_n = 3n^2 + 5n$, find the $n$th term and the common difference.

Solution:

1. $a_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]$.
2. $= 3n^2 + 5n - (3n^2 - 6n + 3 + 5n - 5) = 6n + 2$.
3. $a_1 = 8$, $a_2 = 14$, so $d = 6$.

Final Answer: $a_n = 6n + 2$, $d = 6$.

Takeaway: $a_n = S_n - S_{n-1}$ recovers the term formula.

Example 7: Sum of first 20 odd numbers

Find the sum of the first 20 odd natural numbers.

Solution:

1. AP $1, 3, 5, \dots$: $a = 1$, $d = 2$, $n = 20$.
2. $S_{20} = \dfrac{20}{2}[2 + 19(2)] = 10[2 + 38] = 10(40) = 400$.

Final Answer: 400.

Takeaway: The sum of the first $n$ odd numbers is $n^2$ (here $20^2 = 400$).

Example 8: Find $n$ for a given sum (one root)

How many terms of the AP $9, 17, 25, \dots$ must be taken to give a sum of 636?

Solution:

1. $a = 9$, $d = 8$. $\dfrac{n}{2}[18 + (n-1)8] = 636$.
2. $n(8n + 10) = 1272 \Rightarrow 8n^2 + 10n - 1272 = 0 \Rightarrow 4n^2 + 5n - 636 = 0$.
3. $(4n + 53)(n - 12) = 0 \Rightarrow n = 12$ (reject the negative root).

Final Answer: 12 terms.

Takeaway: Reject the negative/fractional root; only $n = 12$ is admissible.

Example 9: Sum given two conditions

If the sum of the first 14 terms of an AP is 1050 and the first term is 10, find the 20th term.

Solution:

1. $S_{14} = \dfrac{14}{2}[2(10) + 13d] = 7[20 + 13d] = 1050 \Rightarrow 20 + 13d = 150 \Rightarrow d = 10$.
2. $a_{20} = 10 + 19(10) = 200$.

Final Answer: $a_{20} = 200$.

Takeaway: Use the sum to find $d$, then the nth-term formula.

Example 10: Sum of multiples

Find the sum of all multiples of 7 lying between 1 and 100.

Solution:

1. Multiples: $7, 14, \dots, 98$. Here $a = 7$, $d = 7$, $l = 98$.
2. Number of terms: $98 = 7 + (n-1)7 \Rightarrow n = 14$.
3. $S_{14} = \dfrac{14}{2}(7 + 98) = 7 \times 105 = 735$.

Final Answer: 735.

Takeaway: First find $n$, then apply $S_n = \dfrac{n}{2}(a + l)$.