nth term of an A.P.

Let a represent the first term and d represents the common difference of an A.P. Then the n^th term or general term of an A.P. is,

a_n = a + (n − 1) d

OR

n^th term of an A.P. = First term + (Term’s Number − 1) × Common difference

Example: Find 10^th term of an A.P. −40, −15, 10, 35, ….

Solution:

Here, a = −40 and n = 10

Now, determine the common difference d.

    d = −15−(−40)

⇒ d = 25

Substitute −40 for a, 10 for n and 25 for d in a_n = a + (n − 1) d.

     a₁₀ = −40 + (10 − 1) 25

⇒  a₁₀ = −40 + (9)25

⇒  a₁₀ = 185

Therefore, 10^th term of an A.P. is 185.

n^th term of an A.P from the end

Let l represent the last term and d represent the common difference of an A.P. then the n^th term of an A.P. from the end is given as:

n^th term from the end = l − (n − 1) d

n^th term from the end = Last term + (Term number − 1) × Common difference

Example: Find 12^th term from the end of an A.P. 3, 5, 7, 9, …, 201.

Solution:

Here, l = 201 and n = 12

Now, determine the common difference d.

    d = 5 – 3

⇒ d = 2

Substitute 201 for l, 12 for n and 2 for d in n^th term from the end = l − (n − 1) d.

        12^th term from the end = 201 − (12 − 1) 2

 ⇒   12th term from the end = 201 −(11)2

⇒   12th term from the end = 179

Therefore, 12^th term from the end of an A.P. is 179.

Middle term of a finite A.P.

Let a represent the first term, d represents the common difference and n represents the number of terms of a finite A.P.

If n is odd, then ((n + 1) / 2)^th term is the middle term of a finite A.P.

If n is odd, then (n / 2)^th and ((n / 2) + 1)^th terms are the middle term of a finite A.P.

 and 

Example: Find the middle term of an A.P. 213, 205, 197, …, 37.

Solution:

Here, a = 213 and a_n = 37

Now, determine the common difference d.

    d = 205 − 213

⇒ d = −8

Substitute 213 for a, 37 for a_n and −8 for d in a_n = a + (n − 1) d.

    37 = 213 + (n − 1) −8

⇒ 37 = 213 − 8n + 8

⇒ 8n = 213 + 8 – 37

⇒ 8n = 184

⇒   n = 23

Here, n is odd.

So, ((n + 1) / 2)^th term is the middle term of a finite A.P.

Substitute 213 for a, 23 for n and −8 for d in a_(n+1)/2 = a + (((n + 1) / 2) − 1) d.

   a_(23+1)/2 = 213 + (((23 + 1) / 2) − 1) −8

⇒       a₁₂ = 213+ (12 − 1) −8

⇒       a₁₂ = 213 − 88

⇒       a₁₂ = 125

Therefore, the middle term of an A.P. is 125.

CBSE Class 10 Maths Notes Chapter 4 Arithmetic Progressions are an outstanding resource created by our team of knowledgeable Subject Experts at GfG. As ardent supporters of students’ education, we place a high priority on their learning and development, which is why we have written these in-depth notes to aid them in comprehending the challenging subject of arithmetic progressions.

Chapter 4 of the NCERT Class 10 Maths textbook finds the nth term of an arithmetic progression, summing the n terms of an arithmetic progression, calculating the arithmetic mean, and many other topics covered. These notes are intended to give students a thorough overview of the entire chapter, covering all the crucial topics, formulas, and ideas they will need to know to ace their examinations.