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 $0,3,10,21,36.$