Arithmetic Progressions • Topic 4 of 5

Sum of First n Terms of an AP

What is Summation? The sum of the first $ n $ terms of an AP is denoted by $ S_n $. It is the total when you add all terms from the first to the nth term.

Formula for Sum of First n Terms: There are two common formulas: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] \[ S_n = \frac{n}{2} (a + l) \] where $ l = a_n = a + (n-1)d $ is the last term.

Which formula to use?

Use $ S_n = \frac{n}{2}[2a + (n-1)d] $ when you know $ a, d, n $.
Use $ S_n = \frac{n}{2}(a + l) $ when you know $ a, l, n $.

Sum of Finite AP: For a finite AP with $ n $ terms, the sum is always a finite number.

Real-life applications:

Total savings after $ n $ months with fixed monthly increase
Total number of seats in an auditorium with increasing rows
Distance traveled in $ n $ seconds with constant acceleration
Total production over several years with fixed annual increase

SUM OF FIRST n TERMS (Visual Representation)
   AP: 2, 5, 8, 11, 14 (a=2, d=3, n=5)
   
   Write terms forward:   2  +  5  +  8  + 11  + 14 = 40
   Write terms backward: 14 + 11  +  8  +  5  +  2 = 40
   Add column by column: 16 + 16  + 16  + 16  + 16 = 80
   
   Each pair sum = a + l = 2 + 14 = 16
   Number of pairs = n = 5
   Total of both rows = n × (a + l) = 5 × 16 = 80
   So S_n = 80 ÷ 2 = 40 ✓

APPLICATION: SEATING ARRANGEMENT
   Row 1: 10 seats
   Row 2: 13 seats
   Row 3: 16 seats
   Total seats in 8 rows:
   a=10, d=3, n=8
   S₈ = 8/2[2×10 + (8-1)×3]
      = 4[20 + 21] = 4×41 = 164 seats

SUM FORMULAE QUICK REFERENCE
   ┌────────────────────────────────────────┐
   │ S_n = n/2 [2a + (n-1)d]                │
   │ S_n = n/2 (a + l) where l = last term  │
   └────────────────────────────────────────┘

Solved Examples

Worked Example

Find the sum of the first 20 terms of the AP: 5, 11, 17, 23, …

Solution

$ a = 5, d = 6, n = 20 $
$ S_n = \frac{n}{2}[2a + (n-1)d] $
$ S_{20} = \frac{20}{2}[2×5 + (20-1)×6] $
$ = 10[10 + 19×6] = 10[10 + 114] = 10×124 = 1240 $

Answer: The sum is 1240.

Worked Example

How many terms of the AP: 24, 21, 18, … must be taken so that their sum is 78?

Solution

$ a = 24, d = 21-24 = -3, S_n = 78 $
$ \frac{n}{2}[2×24 + (n-1)(-3)] = 78 $
$ \frac{n}{2}[48 - 3n + 3] = 78 $
$ \frac{n}{2}[51 - 3n] = 78 $
Multiply by 2: $ n(51 - 3n) = 156 $
$ 51n - 3n^2 = 156 $ → $ 3n^2 - 51n + 156 = 0 $
Divide by 3: $ n^2 - 17n + 52 = 0 $
$ (n-4)(n-13) = 0 $ → $ n = 4 $ or $ n = 13 $
Both are valid. For $ n=4 $: 24+21+18+15=78; For $ n=13 $: sum is also 78 (terms become negative later)

Answer: 4 terms or 13 terms.

Worked Example

A contract on construction job specifies a penalty for delay beyond a certain date: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, and so on. Find the total penalty if the work is delayed by 30 days.

Solution

Penalties form AP: 200, 250, 300, …
$ a = 200, d = 50, n = 30 $
$ S_{30} = \frac{30}{2}[2×200 + (30-1)×50] $
$ = 15[400 + 29×50] = 15[400 + 1450] $
$ = 15 × 1850 = 27750 $

Answer: Total penalty = ₹27,750.

Key Points

Sum of first $ n $ terms: $ S_n = \frac{n}{2}[2a + (n-1)d] $ or $ S_n = \frac{n}{2}(a + l) $
The second formula is useful when the last term is given.
Sum is finite for a finite number of terms.
Real-life uses include total savings, seating capacity, penalties, and production totals.
To find $ n $ given $ S_n $, form and solve a quadratic equation.

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: $\tfrac{10}{2}(2+20)=110$.

Practice more MCQs → 📝 Assignment · 35 marks