Sum of First n Terms of an Arithmetic Progression

The sum of the first *n* terms of an arithmetic sequence, denoted by $S_n$, is the total when you add all the terms from the first term up to the *n*th term. Understanding how to calculate this sum is crucial in various areas of mathematics and its applications. Arithmetic sequences are sequences where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by *d*. We have two main formulas to easily determine $S_n$. The choice of which formula to use depends on the information you are given.

Formulae

Here are the two key formulas for calculating the sum of the first *n* terms of an arithmetic sequence:

• Formula 1: If you know the first term (*a*), the common difference (*d*), and the number of terms (*n*):
  $S_n = \frac{n}{2}[2a + (n-1)d]$
• Formula 2: If you know the first term (*a*), the last term (*l*), and the number of terms (*n*):
  $S_n = \frac{n}{2}[a + l]$

Examples

Let’s look at some examples to illustrate how to use these formulas.

Example-1: Find the sum of the first 10 terms of the arithmetic sequence: 2, 5, 8, 11,…

Here, the first term (*a*) = 2, the common difference (*d*) = 3, and the number of terms (*n*) = 10.

Using Formula 1:

$S_{10} = \frac{10}{2}[2(2) + (10-1)3] = 5[4 + 27] = 5[31] = 155$

Therefore, the sum of the first 10 terms is 155.

Example-2: Find the sum of the arithmetic sequence: 3, 7, 11, …, 31. (Hint: find *n* first!)

Here, *a* = 3, *l* = 31, and *d* = 4. First, find *n* using the formula for the *n*th term: $l = a + (n-1)d$.

So, $31 = 3 + (n-1)4$. This gives us $28 = (n-1)4$, or $n-1=7$ and therefore $n=8$.

Now, using Formula 2:

$S_8 = \frac{8}{2}[3 + 31] = 4[34] = 136$

Therefore, the sum of the arithmetic sequence is 136.

Theorem with Proof

Theorem: The sum of the first *n* terms of an arithmetic sequence with first term *a* and common difference *d* is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.

Proof:

Let the arithmetic sequence be denoted as: $a, a+d, a+2d, …, a+(n-1)d$.

The sum, $S_n$, can be written as:

$S_n = a + (a+d) + (a+2d) + … + (a+(n-2)d) + (a+(n-1)d)$ (Equation 1)

Now, write the sequence in reverse order:

$S_n = (a+(n-1)d) + (a+(n-2)d) + … + (a+d) + a$ (Equation 2)

Add Equation 1 and Equation 2, term by term:

$2S_n = [a + (a+(n-1)d)] + [(a+d) + (a+(n-2)d)] + … + [(a+(n-2)d) + (a+d)] + [(a+(n-1)d) + a]$

Notice that each term in the brackets sums to $2a + (n-1)d$. Since there are *n* terms in the original sequence, there will be *n* such bracketed terms:

$2S_n = n[2a + (n-1)d]$

Divide both sides by 2:

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

This proves the formula.

Common mistakes by students

• Confusing the formulas: Using the wrong formula based on the information provided. Always identify *a*, *d*, *n*, and *l* clearly before starting.
• Incorrectly identifying *d*: Forgetting to subtract the first term from the *second* term to find *d*. The common difference is found by subtracting consecutive terms.
• Arithmetic errors: Making mistakes in the arithmetic calculations, particularly when multiplying or simplifying expressions with negative numbers. Double-check your calculations.
• Incorrectly finding *n*: When using the second formula ($S_n = \frac{n}{2}[a + l]$), if *l* is given, it’s vital to find *n* correctly using $l = a + (n-1)d$ first.

Real Life Application

Arithmetic sequences and their sums have several real-world applications:

• Salary Increases: If a person’s salary increases by a fixed amount each year, their total earnings over a period of years can be calculated using the sum of an arithmetic series.
• Stacking Objects: Consider stacking logs or bricks where each row has a fixed number less than the row below. The total number of logs or bricks can be found using the sum of an arithmetic series.
• Depreciation: The value of an asset (like a car) depreciates by a fixed amount each year. The total depreciation over a period can be modeled with an arithmetic series.
• Investment with fixed deposits: The interest earned on fixed deposits, where the principal remains the same but the interest earned is added every year.

Fun Fact

The story goes that the mathematician Carl Friedrich Gauss, as a child, was tasked by his teacher to add all the numbers from 1 to 100. Instead of adding them one by one, Gauss quickly realized he could pair the numbers (1+100, 2+99, 3+98, etc.) and each pair would sum to 101. There were 50 such pairs, so the total sum was 50 x 101 = 5050. This shows the power of understanding arithmetic series!

Recommended YouTube Videos for Deeper Understanding

Q.1 What is the sum of the first 10 terms of an arithmetic sequence with first term $a = 3$ and common difference $d = 2$?

Check Solution

Ans: A

Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$, we have $S_{10} = \frac{10}{2}[2(3) + (10-1)2] = 5[6 + 18] = 5(24) = 120$.

Q.2 The sum of the first $n$ terms of an arithmetic sequence is given by $S_n = 3n^2 + 2n$. What is the first term of the sequence?

Check Solution

Ans: C

The first term, $a$, is equal to $S_1$. Therefore, $a = S_1 = 3(1)^2 + 2(1) = 3 + 2 = 5$.

Q.3 An arithmetic sequence has a first term of 5 and a last term of 20. If the sum of the sequence is 125, how many terms are in the sequence?

Check Solution

Ans: B

Use the formula $S_n = \frac{n}{2}[a + l]$. We have $125 = \frac{n}{2}[5 + 20]$, so $125 = \frac{n}{2}(25)$. Then $n = \frac{2(125)}{25} = 10$.

Q.4 If the sum of the first 8 terms of an arithmetic sequence is 100, and the common difference is 3, what is the first term?

Check Solution

Ans: C

Using the formula $S_n = \frac{n}{2}[2a + (n-1)d]$, we have $100 = \frac{8}{2}[2a + (8-1)3]$. Then $100 = 4[2a + 21]$, so $25 = 2a + 21$, which gives $2a = 4$ and $a = 2$.

Q.5 The third term of an arithmetic sequence is 7, and the seventh term is 15. What is the sum of the first 10 terms of this sequence?

Check Solution

Ans: A

We have $a_3 = a + 2d = 7$ and $a_7 = a + 6d = 15$. Subtracting the first equation from the second gives $4d = 8$, so $d = 2$. Substituting into the first equation gives $a + 2(2) = 7$, so $a = 3$. Then $S_{10} = \frac{10}{2}[2(3) + (10-1)2] = 5[6 + 18] = 5(24) = 120$.

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