Arithmetic Progressions • Topic 4 of 5

Sum of First n Terms of an AP

What is $S_n$? The sum of the first $n$ terms of an AP is written $S_n$. It adds up everything from the first term to the $n$th term: $S_n = a_1 + a_2 + a_3 + \dots + a_n$. Adding the terms one by one is slow, so we use a formula.

The pairing idea (how the formula arises). Write the sum forwards and again backwards underneath, then add column by column. Each column sums to (first $+$ last), there are $n$ such columns, and we counted everything twice — hence $S_n = \dfrac{n}{2}(a + l)$. This pairing trick (often credited to a young Gauss) is the reason the $\dfrac{n}{2}$ appears.

The two standard formulas. $S_n = \dfrac{n}{2}\,[2a + (n - 1)d]$ and $S_n = \dfrac{n}{2}\,(a + l)$, where $l = a_n = a + (n - 1)d$ is the last term. The two are the same formula with $l$ written out, so pick whichever fits the data.

  • Use $S_n = \dfrac{n}{2}[2a + (n - 1)d]$ when you know $a$, $d$ and $n$.
  • Use $S_n = \dfrac{n}{2}(a + l)$ when you know $a$, the last term $l$ and $n$.

Recovering a term from the sums. A very useful relation links the sum and the terms: $a_n = S_n - S_{n - 1}$. The $n$th term is whatever the total gains when you include one more term. In particular $a_1 = S_1$. This is the standard way to find a term (or $d$) when only a formula for $S_n$ is given, e.g. $S_n = 3n^2 + 5n$.

Finding $n$ from a given sum. Setting $S_n$ equal to a target value gives an equation in $n$. Because $S_n$ is quadratic in $n$, this is a quadratic equation. Solve it, then keep only positive-integer roots. Sometimes two values of $n$ work — this happens in a decreasing AP, where later negative terms cancel earlier ones to give the same running total twice.

Reading the answer sensibly. $n$ must be a positive whole number. Discard fractions or negatives. If both quadratic roots are valid integers, both are genuine answers and should be reported. If neither is a positive integer, no such $n$ exists.

Where sums show up. Total savings after $n$ months under a fixed-increment plan; total seats across rows that grow steadily; total penalty when the daily fine rises; total logs in a triangular stack; total prize money split into steadily decreasing awards; total distance under uniform acceleration. Each of these is "add up an AP", i.e. a single $S_n$ calculation.

  • Total savings after $n$ months with a fixed monthly increase.
  • Total seats in a hall whose rows grow by a constant.
  • Total penalty when the daily fine rises by a fixed amount.
  • Total logs in a stack where each row has one fewer log.

Common mistakes to avoid. (1) Forgetting the factor $\dfrac{n}{2}$ or dividing only one bracket. (2) Mixing the two formulas — do not put $l$ inside the $[2a + (n - 1)d]$ bracket. (3) Sign slips when $d < 0$. (4) Keeping a non-integer or negative root for $n$. (5) In $a_n = S_n - S_{n - 1}$, miscomputing $S_{n-1}$ by failing to replace every $n$ with $(n - 1)$.

Gauss pairing trick: the AP 2, 5, 8, 11, 14 written forwards and backwards so each column adds to first plus lastThe pairing trick: write forwards, then backwards258111414118521616161616Each column = a + l = 2 + 14 = 16, and there are n = 5 columns2 × S = n(a + l) = 5 × 16 = 80, so S = 40Sum counted twice, hence the ½ factorThe two sum formulas and a staircase showing rows of seats 10, 13, 16 growing by 3Sum formulas and a real applicationTwo standard formulasSn = (n/2)[2a + (n − 1)d]Sn = (n/2)(a + l)use the second when l is known(l = last term)Seats per row (AP)Row 1: 10Row 2: 13Row 3: 16a = 10, d = 3 → S gives total seats
Solved Examples
1
Worked Example
Find the sum of the first 20 terms of the AP 5, 11, 17, 23, ...
Solution
  1. $a = 5$, $d = 6$, $n = 20$.
  2. $S_n = \dfrac{n}{2}[2a + (n - 1)d]$.
  3. $S_{20} = \dfrac{20}{2}[2 \times 5 + 19 \times 6] = 10[10 + 114]$.
  4. $= 10 \times 124 = 1240$.

Answer: The sum is 1240.

2
Worked Example
Find the sum of the first 15 terms of the AP 8, 3, -2, -7, ...
Solution
  1. $a = 8$, $d = -5$, $n = 15$.
  2. $S_{15} = \dfrac{15}{2}[2 \times 8 + 14 \times (-5)]$.
  3. $= \dfrac{15}{2}[16 - 70] = \dfrac{15}{2}(-54)$.
  4. $= 15 \times (-27) = -405$.

Answer: The sum is $-405$.

3
Worked Example
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.
Solution
  1. $a = 5$, $l = 45$, $S_n = 400$.
  2. $S_n = \dfrac{n}{2}(a + l) \Rightarrow 400 = \dfrac{n}{2}(5 + 45) = 25n$.
  3. $n = 16$.
  4. Now $l = a + (n - 1)d \Rightarrow 45 = 5 + 15d \Rightarrow 15d = 40 \Rightarrow d = \dfrac{8}{3}$.

Answer: $n = 16$ and $d = \dfrac{8}{3}$.

4
Worked Example
How many terms of the AP 24, 21, 18, ... must be taken so that their sum is 78?
Solution
  1. $a = 24$, $d = -3$, $S_n = 78$.
  2. $\dfrac{n}{2}[48 + (n - 1)(-3)] = 78 \Rightarrow \dfrac{n}{2}[51 - 3n] = 78$.
  3. $n(51 - 3n) = 156 \Rightarrow 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0$.
  4. $(n - 4)(n - 13) = 0 \Rightarrow n = 4$ or $n = 13$.
  5. Both are positive integers; for $n = 13$ the later negative terms cancel back to 78.

Answer: 4 terms or 13 terms.

5
Worked Example
Find the sum of all two-digit numbers divisible by 3.
Solution
  1. They are $12, 15, 18, \dots, 99$ with $a = 12$, $d = 3$, $l = 99$.
  2. Number of terms: $99 = 12 + (n - 1)3 \Rightarrow (n - 1)3 = 87 \Rightarrow n = 30$.
  3. $S_{30} = \dfrac{30}{2}(12 + 99) = 15 \times 111$.
  4. $= 1665$.

Answer: The sum is 1665.

6
Worked Example
The sum of the first $n$ terms of an AP is $S_n = 3n^2 + 5n$. Find its $n$th term and the 25th term.
Solution
  1. $a_n = S_n - S_{n - 1}$.
  2. $S_{n - 1} = 3(n - 1)^2 + 5(n - 1) = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2$.
  3. $a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2$.
  4. $a_{25} = 6 \times 25 + 2 = 152$.

Answer: $a_n = 6n + 2$; the 25th term is 152.

7
Worked Example
A construction contract has a penalty for delay: ₹200 for the first day, ₹250 for the second, ₹300 for the third, and so on. Find the total penalty if work is delayed by 30 days.
Solution
  1. Penalties: 200, 250, 300, ... — an AP with $a = 200$, $d = 50$.
  2. $S_{30} = \dfrac{30}{2}[2 \times 200 + 29 \times 50]$.
  3. $= 15[400 + 1450] = 15 \times 1850$.
  4. $= 27750$.

Answer: Total penalty $=$ ₹27,750.

8
Worked Example
A sum of ₹700 is to be used for seven cash prizes. Each prize is ₹20 less than its preceding prize. Find the value of each prize.
Solution
  1. Let the largest prize be $a$; the prizes are $a, a - 20, a - 40, \dots$ ($d = -20$, $n = 7$).
  2. $S_7 = \dfrac{7}{2}[2a + 6(-20)] = 700 \Rightarrow \dfrac{7}{2}[2a - 120] = 700$.
  3. $7(a - 60) = 700 \Rightarrow a - 60 = 100 \Rightarrow a = 160$.
  4. Prizes: 160, 140, 120, 100, 80, 60, 40.

Answer: The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.

9
Worked Example
200 logs are stacked so that the bottom row has 20 logs, the next 19, the next 18, and so on. In how many rows are the 200 logs placed, and how many are in the top row?
Solution
  1. Rows have 20, 19, 18, ... logs — an AP with $a = 20$, $d = -1$, total $S_n = 200$.
  2. $\dfrac{n}{2}[40 + (n - 1)(-1)] = 200 \Rightarrow n(41 - n) = 400$.
  3. $n^2 - 41n + 400 = 0 \Rightarrow (n - 16)(n - 25) = 0 \Rightarrow n = 16$ or $25$.
  4. $n = 25$ gives a 25th row of $20 - 24 = -4$ logs (impossible), so $n = 16$.
  5. Top row: $a_{16} = 20 + 15(-1) = 5$ logs.

Answer: 16 rows, with 5 logs in the top row.

10
Worked Example
Find the sum of the first 40 positive integers divisible by 6.
Solution
  1. They are $6, 12, 18, \dots$ — an AP with $a = 6$, $d = 6$, $n = 40$.
  2. $S_{40} = \dfrac{40}{2}[2 \times 6 + 39 \times 6]$.
  3. $= 20[12 + 234] = 20 \times 246$.
  4. $= 4920$.

Answer: The sum is 4920.

11
Worked Example
In an AP the sum of the first 7 terms is 49 and the sum of the first 17 terms is 289. Find the sum of its first $n$ terms.
Solution
  1. $S_7 = \dfrac{7}{2}[2a + 6d] = 49 \Rightarrow 2a + 6d = 14 \Rightarrow a + 3d = 7$ ... (1)
  2. $S_{17} = \dfrac{17}{2}[2a + 16d] = 289 \Rightarrow 2a + 16d = 34 \Rightarrow a + 8d = 17$ ... (2)
  3. Subtract (1) from (2): $5d = 10 \Rightarrow d = 2$; then $a = 1$.
  4. $S_n = \dfrac{n}{2}[2(1) + (n - 1)2] = \dfrac{n}{2}[2n] = n^2$.

Answer: $S_n = n^2$.

12
Worked Example
A man repays a loan of ₹3250 by paying ₹20 in the first month and increasing the payment by ₹15 every month. How long will it take to clear the loan?
Solution
  1. Payments form an AP: $a = 20$, $d = 15$; total $S_n = 3250$.
  2. $\dfrac{n}{2}[2 \times 20 + (n - 1)15] = 3250 \Rightarrow n[40 + 15n - 15] = 6500$.
  3. $15n^2 + 25n - 6500 = 0 \Rightarrow 3n^2 + 5n - 1300 = 0$.
  4. $n = \dfrac{-5 + \sqrt{25 + 15600}}{6} = \dfrac{-5 + 125}{6} = 20$ (reject the negative root).

Answer: The loan is cleared in 20 months.

Key Points

  • Sum of the first $n$ terms: $S_n = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}(a + l)$.
  • Use $a, l, n$ form when the last term is known; use $a, d, n$ form otherwise.
  • Term from sums: $a_n = S_n - S_{n - 1}$, with $a_1 = S_1$.
  • Setting $S_n$ to a target gives a quadratic in $n$ — keep only positive-integer roots.
  • Two valid values of $n$ can occur in a decreasing AP, where negative terms cancel earlier ones.
  • Word problems on savings, penalties, prizes, stacking and loan repayment are all single $S_n$ calculations.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The sum of $n$ terms of an AP is:
Explanation: Standard sum formula.
Q2.Also $S_n=\tfrac{n}{2}(a+l)$, where $l$ is the:
Explanation: $l$ is the last term.
Q3.$1+2+\dots+n=$
Explanation: Sum of first $n$ naturals.
Q4.The sum of the first $10$ terms of $2,4,6,\dots$ is:
Explanation: $\tfrac{10}{2}(2+20)=110$.
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