Login/Sign Up
Home » 1. Write the first terms of each of the following sequences whose term are: (vii) (viii)
1. Write the first terms of each of the following sequences whose {{n}^{th}} term are: (vii) {{a}_{n}}={{n}^{2}}-n+1 (viii) {{a}_{n}}={{n}^{2}}-n+1
Arithmetic Progression | Class 10 | ICSE | Maths | RD Sharma
1. Write the first terms of each of the following sequences whose {{n}^{th}} term are: (vii) {{a}_{n}}={{n}^{2}}-n+1 (viii) {{a}_{n}}={{n}^{2}}-n+1

An arithmetic progressions or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solutions:

(vii) {{a}_{n}}={{n}^{2}}-n+1

The given sequence whose, {{a}_{n}}={{n}^{2}}-n+1

To get the first five terms of given sequence, put n=1,2,3,4,5.

And, we get

{{a}_{1}}={{1}^{2}}-1+1=1

{{a}_{2}}={{2}^{2}}-2+1=3

{{a}_{3}}={{3}^{2}}-3+1=7

{{a}_{4}}={{4}^{2}}-4+1=13

{{a}_{5}}={{5}^{2}}-5+1=21

\therefore the required first five terms of the sequence are 1,3,7,13,21.

(viii) {{a}_{n}}=2{{n}^{2}}-3n+1

The given sequence whose {{a}_{n}}=2{{n}^{2}}-3n+1

To get the first five terms of the sequence, put n=1,2,3,4,5.

And, we get

{{a}_{1}}={{2.1}^{2}}-3.1+1=2-3+1=0

{{a}_{2}}={{2.2}^{2}}-3.2+1=8-6+1=3

{{a}_{3}}={{2.3}^{2}}-3.3+1=18-9+1=10

{{a}_{4}}={{2.4}^{2}}-3.4+1=32-12+1=21

{{a}_{5}}={{2.5}^{2}}-3.5+1=50-15+1=36

\therefore the required first five terms of the sequence are 0,3,10,21,36.

i

More Material