CBSE Class 10 Mathematics Arithmetic Progressions Notes Set 11

Arithmetic Progressions

A sequence in which the difference between each term and the one before it is the same constant is an arithmetic progression. That constant is the common difference d. With first term a the terms are a, a + d, a + 2d, and so on. If the nth term is linear in n, such as \( an+b \), the sequence is an A.P. with common difference a.
Adding or subtracting a number to every term, or multiplying every term by a non-zero constant k, keeps the sequence an A.P. (the difference becomes kd). Three numbers a, b, c are in A.P. when \( 2b=a+c \).

The nth Term

\( a_n=a+(n-1)d \). To test whether a number is a term, solve for n; it is a term only if n is a natural number. For the first negative term solve \( a+(n-1)d<0 \) and take the least integer n. The mth term from the end of an A.P. with n terms is the \( (n-m+1) \)th term from the beginning.

Choosing Terms Cleverly

For three terms use \( a-d,\ a,\ a+d \); for five terms \( a-2d,\ a-d,\ a,\ a+d,\ a+2d \); for four terms \( a-3d,\ a-d,\ a+d,\ a+3d \) with common difference 2d. The sum of the terms then simplifies and only d appears squared.

Sum of n Terms

\( S_n=\frac n2\{2a+(n-1)d\}=\frac n2(a+l) \), where l is the last term. Also \( a_n=S_n-S_{n-1} \) for n of 2 or more. When two answers appear for n (for example if some terms are zero or negative), check which is meaningful in the story.
Examples: the sum of the first 20 terms of 1, 4, 7, ... is 590; the three-digit multiples of 7 run from 105 to 994, so there are 128 of them and their sum is 70336; if \( S_n=3n^2-4n \) then the A.P. is -1, 5, 11, ... and \( a_{12}=65 \).