Sum of n Terms

Finding the sum of n terms is one of the key ideas in mathematics that is very important for all students, professionals etc., who deal with ever-increasing amounts or patterns or series. It is very important to have an understanding of how to find the sum of many things (n terms) in a sequence. Whenever you are working with payments, examining scientific experiments or figuring out a solution to a math problem, understanding how to find the total amount of many items is an important skill to develop. 

In this guide we will go through all the different formulas, derivations, applications, and common errors made when looking at the sum of n terms, paying close attention to APs, GPs, and other sequences alike.

1.0What Does the "Sum of First n Terms" Mean?

The arithmetic sequence or arithmetic progressions is a series of numbers. You get more numbers by adding the same number every time you take one number from your arithmetic sequence or arithmetic progression.

The standard format of an AP looks like this:

a, \ a+d, \ a+2d, \ a+3d, \ \dots

What we mean by Finding the sum of First n Terms i.e. (S) or (S_n) is the total amount of numbers that is created by adding together all of the Numbers between the First Term and the (n)th Number.

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

2.0

3.0Derivation of the AP Sum Formula

Let us formally derive the mathematical formula using the textbook method of reversing the terms.

Let S represent the total sum of the first $n$ terms. We can write our series normally as:

S = a + (a + d) + (a + 2d) + \dots + [a + (n - 2)d] + [a + (n - 1)d] \quad \text{--- (Equation 1)}

Now, let us write down the exact same sum, but completely reverse the order of the terms, starting from the last term and working backwards to the first:

S = [a + (n - 1)d] + [a + (n - 2)d] + \dots + (a + d) + a \quad \text{--- (Equation 2)}

Now, we add Equation 1 and Equation 2 together term-wise (adding the first terms together, the second terms together, and so on):

• First vertical pair: $a + [a + (n - 1)d] = 2a + (n - 1)d
• Second vertical pair: $(a + d) + [a + (n - 2)d] = 2a + d + nd - 2d = 2a + (n - 1)d
• Last vertical pair: $[a + (n - 1)d] + a = 2a + (n - 1)d

Notice a remarkable consistency: every single paired column adds up to the exact same value, which is 2a + (n - 1)d.

Since our sequence contains exactly $n$ terms, we will have exactly $n$ of these identical combinations on the right-hand side:

S + S = [2a + (n - 1)d] + [2a + (n - 1)d] + \dots + [2a + (n - 1)d] \quad \text{(n times)}

2S = n \cdot [2a + (n - 1)d]

Dividing both sides by 2 gives us our foundational formula:

The Sum of n Terms Formula (Standard Form):

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

The Alternate Form: Connecting to the Last Term ($l$)

We can simplify this formula even further depending on what information is provided in a problem. Let us take a look inside the brackets of our standard formula and split the 2a term into a + a

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

If you recall the formula for the general n-th term of an AP, it is written as a_n = a + (n - 1)d. If a finite AP has exactly $n$ terms, this $n$-th term is also called the last term, denoted by the letter l.

Substituting l = a + (n - 1)d back into our split equation yields a compact alternative form:

The Last Term Sum Formula:

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

4.0Step-by-Step Solved Examples

Let us cement your understanding by working through four unique problem variations.

Example 1: Finding the sum when a, d, and n are directly given

Question: Find the sum of the first 20 terms of the AP: 3, 8, 13, 18 …….

Solution:

• Step 1: Extract the known parameters from the sequence.
• • First term ($a$) = $3$
  • Common difference ($d$) = $8 - 3 = 5$
  • Number of terms ($n$) = $20$
• Step 2: Write down the standard sum formula.
  $$S = \frac{n}{2} [2a + (n - 1)d]$$
• Step 3: Substitute the values and evaluate step-by-step.
  $$S_{20} = \frac{20}{2} [2(3) + (20 - 1)5]$$
  $$S_{20} = 10 \cdot [6 + (19 \times 5)]$$
  $$S_{20} = 10 \cdot [6 + 95]$$
  $$S_{20} = 10 \cdot 101 = 1010$$

Answer: The sum of the first 20 terms of the given AP is 1,010.

Example 2: Finding the sum of a series when the last term is known

Question: Find the total sum of the arithmetic series: 5 + 12 + 19 + \dots + 145.

Solution:

• Step 1: Identify the known components.
• • First term ($a$) = $5$
  • Common difference ($d$) = $12 - 5 = 7$
  • Last term ($l$ or $a_n$) = $145$
• Step 2: We do not know the value of $n$ (the number of terms), so we must find it first using the general term formula.
  $$a_n = a + (n - 1)d$$
  $$145 = 5 + (n - 1)7$$
  $$145 - 5 = 7(n - 1)$$
  $$140 = 7(n - 1) \implies n - 1 = \frac{140}{7} = 20$$
  $$n = 20 + 1 = 21$$
• Step 3: Now use the simpler last-term formula to find the final sum.
  $$S = \frac{n}{2} (a + l)$$
  $$S_{21} = \frac{21}{2} (5 + 145)$$
  $$S_{21} = \frac{21}{2} \cdot 150 = 21 \times 75 = 1575$$

Answer: The sum of the series is 1,575.

Example 3: Finding missing variables given the sum

Question: The first term of an AP is 6 and the last term is 96. If the sum of all these terms is 816, find the total number of terms (n) and the common difference (d).

Solution:

• Step 1: Identify the given values.
• • $a = 6$, $l = 96$, $S = 816$
• Step 2: Use the last term formula to isolate $n$.
  $$S = \frac{n}{2} (a + l)$$
  $$816 = \frac{n}{2} (6 + 96)$$
  $$816 = \frac{n}{2} \cdot 102$$
  $$816 = 51n \implies n = \frac{816}{51} = 16$$
• Step 3: Use the calculated value of $n = 16$ to find the common difference $d$ via the last term expression.
  $$l = a + (n - 1)d$$
  $$96 = 6 + (16 - 1)d$$
  $$96 - 6 = 15d$$
  $$90 = 15d \implies d = \frac{90}{15} = 6$$

Answer: The number of terms ($n$) is 16, and the common difference ($d$) is 6.

Example 4: Real-world word problem

Question: A professional decides to set up a monthly savings plan. They save ₹200 in the first month, ₹250 in the second month, ₹300 in the third month, and continue increasing their savings by ₹50 every month. What will be their total accumulated savings at the end of 2 years?

Solution:

• Step 1: Translate the real-world scenario into an AP.
• • Monthly savings progression: $200, 250, 300, \dots$
  • First term ($a$) = $200$
  • Common difference ($d$) = $50$
  • Time duration = 2 years = $24\text{ months} \implies n = 24$
• Step 2: Apply the standard sum formula.
  $$S = \frac{n}{2} [2a + (n - 1)d]$$
  $$S_{24} = \frac{24}{2} [2(200) + (24 - 1)50]$$
  $$S_{24} = 12 \cdot [400 + (23 \times 50)]$$
  $$S_{24} = 12 \cdot [400 + 1150]$$
  $$S_{24} = 12 \cdot 1550 = 18600$$

Answer: The total accumulated savings at the end of 2 years will be ₹18,600.