To Find the nᵗʰ Term of an Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant difference to the preceding term. This constant is called the common difference and is denoted by \( d \).

The general form of an AP is:

\[ a, a + d, a + 2d, a + 3d, \ldots \]

where \( a \) is the first term.

Formula Derivation

The \( n^{th} \) term of an AP, denoted by \( a_n \), is given by:

\[ a_n = a + (n - 1)d \]

This formula is derived by starting from the first term and adding the common difference \( d \) \( (n-1) \) times.

If the AP is considered from the end, with last term \( l \), the \( n^{th} \) term from the end is:

\[ a_n = l - (n - 1)d \]

Worked Illustrations and Solved Examples

Example 1: Find the 10ᵗʰ term of the AP: 2, 7, 12, ...

Solution:

• First term, \( a = 2 \)
• Common difference, \( d = 7 - 2 = 5 \)
• Number of terms, \( n = 10 \)
• Using the formula:

\[ a_{10} = 2 + (10 - 1) \times 5 = 2 + 45 = 47 \]

Hence, the 10ᵗʰ term is 47.

Example 2: Determine the AP whose 3ʳᵈ term is 5 and 7ᵗʰ term is 9.

Solution:

Given:

\[ a_3 = a + 2d = 5 \]

\[ a_7 = a + 6d = 9 \]

Subtracting the first from the second:

\[ (a + 6d) - (a + 2d) = 9 - 5 \Rightarrow 4d = 4 \Rightarrow d = 1 \]

Substitute \( d = 1 \) into \( a + 2d = 5 \):

\[ a + 2 = 5 \Rightarrow a = 3 \]

Therefore, the AP is: 3, 4, 5, 6, 7, 8, ...

Example 3: Check whether 301 is a term of the AP: 5, 11, 17, 23, ...

Solution:

First term, \( a = 5 \)

Common difference, \( d = 6 \)

Using the formula:

\[ 301 = 5 + (n - 1)6 \Rightarrow 301 - 5 = 6(n - 1) \Rightarrow 296 = 6n - 6 \Rightarrow 6n = 302 \Rightarrow n = \frac{302}{6} = 50.333... \]

Since \( n \) is not an integer, 301 is not a term of the AP.

Example 4: A flower bed has 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there?

Solution:

First term, \( a = 23 \)

Common difference, \( d = 21 - 23 = -2 \)

Last term, \( a_n = 5 \)

Using the formula:

\[ 5 = 23 + (n - 1)(-2) \Rightarrow 5 = 23 - 2n + 2 \Rightarrow 5 = 25 - 2n \Rightarrow -20 = -2n \Rightarrow n = 10 \]

There are 10 rows in the flower bed.

Practice Set

• Level 1 – Easy
• Find the 15ᵗʰ term of the AP: 3, 6, 9, ...
• Find the 7ᵗʰ term of the AP: 10, 7, 4, ...
• Level 2 – Moderate
• Determine the AP if the 5ᵗʰ term is 20 and the 8ᵗʰ term is 32.
• Check if 50 is a term of the AP: 4, 9, 14, 19, ...
• Level 3 – Challenging
• Find the number of terms in the AP: 100, 95, 90, ... that ends with 10.
• Find the 20ᵗʰ term of the AP whose first term is 7 and the sum of first 20 terms is 610.

Answer Key

• Level 1
• \( a_{15} = 3 + (15 - 1)3 = 3 + 42 = 45 \)
• \( a_7 = 10 + (7 - 1)(-3) = 10 - 18 = -8 \)
• Level 2
• From \( a_5 = a + 4d = 20 \) and \( a_8 = a + 7d = 32 \), subtract to get \( 3d = 12 \Rightarrow d = 4 \). Then \( a + 4 \times 4 = 20 \Rightarrow a = 4 \). AP is 4, 8, 12, ...
• Check \( n \) for 50: \( 50 = 4 + (n - 1)5 \Rightarrow 46 = 5(n - 1) \Rightarrow n = 10.2 \) Not an integer, so 50 is not a term.
• Level 3
• Find \( n \) such that \( a_n = 10 \): \( 10 = 100 + (n - 1)(-5) \Rightarrow -90 = -5n + 5 \Rightarrow -95 = -5n \Rightarrow n = 19 \)
• Sum formula: \( S_n = \frac{n}{2}[2a + (n - 1)d] = 610 \), \( n=20, a=7 \), solve for \( d \):
• \( 610 = 10[14 + 19d] \Rightarrow 61 = 14 + 19d \Rightarrow 19d = 47 \Rightarrow d = \frac{47}{19} = 2.47 \)
• Then \( a_{20} = 7 + 19 \times 2.47 = 7 + 46.93 = 53.93 \)

Quick Reference

Term	Formula	Explanation
nᵗʰ term	\( a_n = a + (n - 1)d \)	Finds the value of the \( n^{th} \) term
nᵗʰ term from end	\( a_n = l - (n - 1)d \)	Finds the \( n^{th} \) term counting backward from last term \( l \)

Glossary

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant.
• Common Difference (d): The fixed amount added to each term to get the next term.
• First Term (a): The initial term of the AP.
• Last Term (l): The final term of a finite AP.
• Term Number (n): The position of a term in the sequence.

Sum of n Terms of an Arithmetic Progression

The sum of the first \( n \) terms of an arithmetic progression is the total obtained by adding all terms from the first to the \( n^{th} \) term.

Formula Derivation

Let the AP be:

\[ a, a + d, a + 2d, \ldots, a + (n - 1)d \]

The sum of the first \( n \) terms, denoted by \( S_n \), is:

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

Writing the sum in reverse order:

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

Adding these two expressions term-wise:

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

Therefore,

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

If the first term \( a \) and last term \( l = a + (n - 1)d \) are known, the sum can also be expressed as:

\[ S_n = \frac{n}{2} (a + l) \]

Worked Illustrations and Solved Examples

Example 5: Find the sum of the first 22 terms of the AP: 8, 3, -2, ...

Solution:

• First term, \( a = 8 \)
• Common difference, \( d = 3 - 8 = -5 \)
• Number of terms, \( n = 22 \)
• Using the formula:

\[ S_{22} = \frac{22}{2} [2 \times 8 + (22 - 1)(-5)] = 11 [16 - 105] = 11 \times (-89) = -979 \]

Hence, the sum is \( -979 \).

Example 6: How many terms of the AP: 24, 21, 18, ... must be taken so that their sum is 78?

Solution:

• First term, \( a = 24 \)
• Common difference, \( d = 21 - 24 = -3 \)
• Sum, \( S_n = 78 \)

Using the sum formula:

\[ 78 = \frac{n}{2} [2 \times 24 + (n - 1)(-3)] \Rightarrow 156 = n [48 - 3n + 3] = n (51 - 3n) \]

Rearranged:

\[ 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0 \]

Factoring:

\[ (n - 4)(n - 13) = 0 \Rightarrow n = 4 \text{ or } 13 \]

Both values are valid.

Example 7: Find the sum of the first 24 terms of the AP whose \( n^{th} \) term is given by \( a_n = 3 + 2n \).

Solution:

Calculate first term:

\[ a_1 = 3 + 2(1) = 5 \]

Calculate second term:

\[ a_2 = 3 + 2(2) = 7 \]

Common difference:

\[ d = 7 - 5 = 2 \]

Number of terms, \( n = 24 \)

Sum:

\[ S_{24} = \frac{24}{2} [2 \times 5 + (24 - 1) \times 2] = 12 [10 + 46] = 12 \times 56 = 672 \]

Hence, the sum is 672.

Practice Set

• Level 1 – Easy
• Find the sum of the first 10 terms of the AP: 1, 3, 5, ...
• Find the sum of the first 15 terms of the AP: 20, 18, 16, ...
• Level 2 – Moderate
• Find the number of terms in the AP: 5, 8, 11, ... if the sum is 195.
• Find the sum of the AP whose first term is 7, common difference is 3, and number of terms is 20.
• Level 3 – Challenging
• Find the sum of the AP: 3, 7, 11, ..., 99.
• Find the sum of the first \( n \) terms of the AP: 2, 5, 8, ... if the sum is 527.

Answer Key

• Level 1
• \( S_{10} = \frac{10}{2} [2 \times 1 + (10 - 1) \times 2] = 5 [2 + 18] = 5 \times 20 = 100 \)
• \( S_{15} = \frac{15}{2} [2 \times 20 + (15 - 1)(-2)] = \frac{15}{2} [40 - 28] = \frac{15}{2} \times 12 = 90 \)
• Level 2
• Sum formula: \( 195 = \frac{n}{2} [2 \times 5 + (n - 1)3] \Rightarrow 390 = n (10 + 3n - 3) = n (7 + 3n) \Rightarrow 3n^2 + 7n - 390 = 0 \)
• Solving quadratic gives \( n = 13 \) (positive root).
• Sum: \( S_{20} = \frac{20}{2} [2 \times 7 + (20 - 1)3] = 10 [14 + 57] = 10 \times 71 = 710 \)
• Level 3
• Number of terms in AP 3 to 99 with \( d=4 \): \( n = \frac{99 - 3}{4} + 1 = 25 \)
• Sum: \( S_{25} = \frac{25}{2} (3 + 99) = \frac{25}{2} \times 102 = 1275 \)
• Sum formula: \( 527 = \frac{n}{2} [2 \times 2 + (n - 1)3] = \frac{n}{2} (4 + 3n - 3) = \frac{n}{2} (1 + 3n) \Rightarrow 1054 = n (1 + 3n) \Rightarrow 3n^2 + n - 1054 = 0 \)
• Solving quadratic gives \( n = 18 \) (positive root).

Quick Reference

Sum	Formula	Explanation
Sum of first n terms	\( S_n = \frac{n}{2} [2a + (n - 1)d] \)	Sum using first term and common difference
Sum using first and last term	\( S_n = \frac{n}{2} (a + l) \)	Sum using first and last term

Glossary

• Sum of n terms (\( S_n \)): The total of the first \( n \) terms of an AP.
• Last term (\( l \)): The \( n^{th} \) term of a finite AP.
• Common difference (\( d \)): The constant difference between consecutive terms.
• Number of terms (\( n \)): The count of terms summed.