Sum of First n Terms of an AP: Formula, Steps, and Examples

The sum of the first $\boldsymbol{n}$ terms of an AP is a widely used formula in mathematics to find the total value of terms in a sequence without manually adding each term. This formula is applicable in solving problems across finance, physics, and engineering.

Formula for the Sum of First n Terms

The sum of the first $\boldsymbol{n}$ terms of an Arithmetic Progression (AP) is denoted by $\boldsymbol{S_n}$ . There are two primary formulas:

If the first term ( $\boldsymbol{a}$ ) and the common difference ( $\boldsymbol{d}$ ) are known:

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

If the first term ( $\boldsymbol{a}$ ) and the last term ( $\boldsymbol{l}$ ) are known:

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

Here:

$\boldsymbol{S_n}$ = the sum of the first $\boldsymbol{n}$ terms,
$\boldsymbol{a}$ = the first term of the AP,
$\boldsymbol{d}$ = the common difference between consecutive terms,
$\boldsymbol{n}$ = the total number of terms,
$\boldsymbol{l}$ = the last term of the AP.

Derivation of the Formula

Using the Sequence Structure

Consider an AP: $\boldsymbol{a, , a + d, , a + 2d, \dots, a + (n - 1)d}$ .
The sum of the first $\boldsymbol{n}$ terms can be expressed as:
$\boldsymbol{S_n = a + (a + d) + (a + 2d) + \dots + (a + (n - 1)d)}$ .

Writing this sum in reverse order:
$\boldsymbol{S_n = (a + (n - 1)d) + (a + (n - 2)d) + \dots + a}$ .

Add these two equations:
$\boldsymbol{2S_n = n \big[2a + (n - 1)d\big]}$ .

Divide both sides by $\boldsymbol{2}$ :
$\displaystyle\boldsymbol{S_n = \frac{n}{2} \big[2a + (n - 1)d\big]}$ .

Examples with Step-by-Step Solutions

Example 1: Find the sum of the first 15 terms of the AP $\boldsymbol{3, 7, 11, 15, \dots}$ .

Solution:

Identify the first term and the common difference:
$\boldsymbol{a = 3, , d = 7 - 3 = 4}$ .
Use the formula for the sum of the first $\boldsymbol{n}$ terms:
$\displaystyle\boldsymbol{S_n = \frac{n}{2} \big[2a + (n - 1)d\big]}$ .
Substitute the values of $\boldsymbol{a}$ , $\boldsymbol{d}$ , and $\boldsymbol{n = 15}$ :
$\displaystyle\boldsymbol{S_{15} = \frac{15}{2} \big[2(3) + (15 - 1)(4)\big]}$ .
Simplify the expression:
$\displaystyle\boldsymbol{S_{15} = \frac{15}{2} \big[6 + 56\big]}$
$\displaystyle\boldsymbol{S_{15} = \frac{15}{2} \cdot 62 = 465}$ .

Answer: The sum of the first 15 terms is $\boldsymbol{S_{15} = 465}$ .

Example 2: The sum of the first 10 terms of an AP is $\boldsymbol{220}$ , and the first term is $\boldsymbol{12}$ . Find the common difference.

Solution:

Use the formula for the sum of the first $\boldsymbol{n}$ terms:
$\displaystyle\boldsymbol{S_n = \frac{n}{2} \big[2a + (n - 1)d\big]}$ .
Substitute the known values:
$\displaystyle\boldsymbol{220 = \frac{10}{2} \big[2(12) + (10 - 1)d\big]}$ .
Simplify the equation:
$\boldsymbol{220 = 5 \big[24 + 9d\big]}$
$\boldsymbol{220 = 120 + 45d}$ .
Solve for $\boldsymbol{d}$ :
$\boldsymbol{220 - 120 = 45d}$
$\boldsymbol{100 = 45d}$
$\displaystyle\boldsymbol{d = \frac{100}{45}}$
$\displaystyle\boldsymbol{d = \frac{20}{9}}$ .

Answer: The common difference is $\displaystyle\boldsymbol{d = \frac{20}{9}}$ .

Example 3: Find the sum of all multiples of $\boldsymbol{3}$ between $\boldsymbol{1}$ and $\boldsymbol{100}$ .

Solution:

Form the AP:
The multiples of $\boldsymbol{3}$ between $\boldsymbol{1}$ and $\boldsymbol{100}$ are:
$\boldsymbol{3, 6, 9, \dots, 99}$ .
Here, $\boldsymbol{a = 3}$ , $\boldsymbol{d = 3}$ , and $\boldsymbol{l = 99}$ .
Find the number of terms ( $\boldsymbol{n}$ ):
Use the formula for the nth term:
$\boldsymbol{a_n = a + (n - 1)d}$ .
Substituting $\boldsymbol{a_n = 99}$ , $\boldsymbol{a = 3}$ , and $\boldsymbol{d = 3}$ :
$\boldsymbol{99 = 3 + (n - 1)(3)}$
$\boldsymbol{96 = 3(n - 1)}$
$\displaystyle\boldsymbol{n - 1 = \frac{96}{3}}$
$\boldsymbol{n = 33}$ .
Use the formula for the sum of the first $\boldsymbol{n}$ terms:
$\displaystyle\boldsymbol{S_n = \frac{n}{2} (a + l)}$ .
Substituting $\boldsymbol{n = 33}$ , $\boldsymbol{a = 3}$ , and $\boldsymbol{l = 99}$ :
$\displaystyle\boldsymbol{S_{33} = \frac{33}{2} (3 + 99)}$
$\displaystyle\boldsymbol{S_{33} = \frac{33}{2} \cdot 102}$
$\displaystyle\boldsymbol{S_{33} = 1683}$ .

Answer: The sum of all multiples of $\boldsymbol{3}$ between $\boldsymbol{1}$ and $\boldsymbol{100}$ is $\boldsymbol{S_{33} = 1683}$ .

Example 4: Find the sum of the first 22 terms of the AP: $\boldsymbol{8, 3, -2, \dots}$ .

Solution:

Identify the values:
$\boldsymbol{a = 8}$ , $\boldsymbol{d = 3 - 8 = -5}$ , $\boldsymbol{n = 22}$ .
Use the formula for the sum of $\boldsymbol{n}$ terms:
$\displaystyle\boldsymbol{S_n = \frac{n}{2} \big[2a + (n - 1)d\big]}$
Substitute the values:
$\displaystyle\boldsymbol{S = \frac{22}{2} \big[2(8) + (22 - 1)(-5)\big]}$
Simplify the expression:
$\boldsymbol{S = 11 \big[16 + 21(-5)\big]}$
$\boldsymbol{S = 11 \big[16 - 105\big]}$
$\boldsymbol{S = 11(-89)}$
$\boldsymbol{S = -979}$

Answer: $\boldsymbol{S = -979}$

Example 5: If the sum of the first 14 terms of an AP is $\boldsymbol{1050}$ , and its first term is $\boldsymbol{10}$ , find the 20th term.

Solution:

Identify the values:
$\boldsymbol{S_{14} = 1050}$ , $\boldsymbol{n = 14}$ , $\boldsymbol{a = 10}$ .
Use the formula for the sum of $\boldsymbol{n}$ terms:
$\displaystyle\boldsymbol{S_n = \frac{n}{2} \big[2a + (n - 1)d\big]}$
Substitute the values for the 14th term sum:
$\displaystyle\boldsymbol{1050 = \frac{14}{2} \big[2(10) + (14 - 1)d\big]}$
Simplify the equation:
$\boldsymbol{1050 = 7 \big[20 + 13d\big]}$
$\boldsymbol{1050 = 140 + 91d}$
$\boldsymbol{910 = 91d}$
$\boldsymbol{d = 10}$
Find the 20th term using the nth term formula:
$\boldsymbol{a_n = a + (n - 1)d}$
$\boldsymbol{a_{20} = 10 + (20 - 1)(10)}$
$\boldsymbol{a_{20} = 10 + 190}$
$\boldsymbol{a_{20} = 200}$

Answer: $\boldsymbol{a_{20} = 200}$

Example 6: Find the sum of the first 24 terms of the list of numbers whose nth term is given by $\boldsymbol{a_n = 3 + 2n}$ .

Solution:

Expand the sequence:
$\boldsymbol{a_1 = 3 + 2(1) = 5}$ , $\boldsymbol{a_2 = 3 + 2(2) = 7}$ , $\boldsymbol{a_3 = 3 + 2(3) = 9}$ , $\dots$
Identify the values:
$\boldsymbol{a = 5}$ , $\boldsymbol{d = 7 - 5 = 2}$ , $\boldsymbol{n = 24}$ .
Use the formula for the sum of $\boldsymbol{n}$ terms:
$\displaystyle\boldsymbol{S_n = \frac{n}{2} \big[2a + (n - 1)d\big]}$
Substitute the values:
$\displaystyle\boldsymbol{S_{24} = \frac{24}{2} \big[2(5) + (24 - 1)(2)\big]}$
Simplify the expression:
$\boldsymbol{S_{24} = 12 \big[10 + 46\big]}$
$\boldsymbol{S_{24} = 12 \cdot 56}$
$\boldsymbol{S_{24} = 672}$

Answer: $\boldsymbol{S_{24} = 672}$

Example 7: A manufacturer produces 600 TV sets in the third year and 700 in the seventh year. Assuming production increases uniformly, find:

The production in the first year,
The production in the 10th year,
The total production in the first 7 years.

Solution:

Since production increases uniformly, it forms an AP. Let the first year production be $\boldsymbol{a}$ and the common difference be $\boldsymbol{d}$ .
From the data:
$\boldsymbol{a + 2d = 600}$
$\boldsymbol{a + 6d = 700}$
Subtract the first equation from the second:
$\boldsymbol{4d = 100 \quad \Rightarrow \quad d = 25}$
Substitute $\boldsymbol{d = 25}$ into $\boldsymbol{a + 2d = 600}$ :
$\boldsymbol{a + 50 = 600 \quad \Rightarrow \quad a = 550}$
Answer: Production in the first year: $\boldsymbol{a = 550}$ .
Find the production in the 10th year:
$\boldsymbol{a_{10} = a + (10 - 1)d}$
$\boldsymbol{a_{10} = 550 + 9(25)}$
$\boldsymbol{a_{10} = 550 + 225}$
$\boldsymbol{a_{10} = 775}$
Answer: Production in the 10th year: $\boldsymbol{a_{10} = 775}$ .
Find the total production in the first 7 years:
$\displaystyle\boldsymbol{S_7 = \frac{7}{2} \big[2a + (7 - 1)d\big]}$
$\displaystyle\boldsymbol{S_7 = \frac{7}{2} \big[2(550) + 6(25)\big]}$
$\displaystyle\boldsymbol{S_7 = \frac{7}{2} \big[1100 + 150\big]}$
$\displaystyle\boldsymbol{S_7 = \frac{7}{2} \cdot 1250}$
$\boldsymbol{S_7 = 4375}$

Answer: Total production in the first 7 years: $\boldsymbol{S_7 = 4375}$

Practice Problems

Find the sum of the first $\boldsymbol{20}$ terms of the AP $\boldsymbol{4, 9, 14, 19, \dots}$ .
Answer: $\boldsymbol{S_{20} = 1240}$ .
The 8th term of an AP is $\boldsymbol{37}$ , and the sum of the first $\boldsymbol{8}$ terms is $\boldsymbol{200}$ . Find the first term.
Answer: $\boldsymbol{a = 13}$ .
Find the sum of all two-digit numbers divisible by $\boldsymbol{7}$ .
Answer: $\boldsymbol{S = 735}$ .

FAQs

How is the sum of an AP derived?admin2024-12-18T14:05:59+05:30

Sum of First n Terms of an AP

Formula for the Sum of First n Terms

Derivation of the Formula

Using the Sequence Structure

Examples with Step-by-Step Solutions

Practice Problems

FAQs

Related Topics

Related Posts

Table of Contents

Join Deeksha Vedantu

> PU + Competitive Exam CoachingPreferred Choice For Toppers25+ Years of Academic Excellence70k+ Success Stories

Related Pages

Latest Posts

Top Links

Top Resources

Top Links

Contact Us

Head Office

Get Social