nth term of an A.P.
Let a represent the first term and d represents the common difference of an A.P. Then the nth term or general term of an A.P. is,
an = a + (n − 1) d
OR
nth term of an A.P. = First term + (Term’s Number − 1) × Common difference
Example: Find 10th 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 an = a + (n − 1) d.
a10 = −40 + (10 − 1) 25
⇒ a10 = −40 + (9)25
⇒ a10 = 185
Therefore, 10th term of an A.P. is 185.
nth 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 nth term of an A.P. from the end is given as:
nth term from the end = l − (n − 1) d
nth term from the end = Last term + (Term number − 1) × Common difference
Example: Find 12th 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 nth term from the end = l − (n − 1) d.
12th term from the end = 201 − (12 − 1) 2
⇒ 12th term from the end = 201 −(11)2
⇒ 12th term from the end = 179
Therefore, 12th 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 an = 37
Now, determine the common difference d.
d = 205 − 213
⇒ d = −8
Substitute 213 for a, 37 for an and −8 for d in an = 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.
Arithmetic Progressions Class 10 Maths Notes Chapter 5
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.