Answer:
[tex]S_{n-1}=0.1n^2 +7.7n - 7.8[/tex]
The diameter of the roll when one sheet is rolled over = 8.2cm
Step-by-step explanation:
(i) Given [tex]S_n = 0.1n^2 + 7.9n[/tex]
To find [tex]S_{n-1}[/tex], substitute the value of ‘n’ as ‘n-1’ in the above equation
Then,
[tex]S_{n-1} = 0.1(n-1)^2 + 7.9(n-1)[/tex]
= [tex]0.1(n^2 -2n+1) +7.9n-7.9[/tex]
=[tex]0.1n^2 -0.2n+0.1 +7.9n-7.9[/tex]
=[tex]0.1n^2 +7.7n - 7.8[/tex]
Hence, [tex]S_{n-1}=0.1n^2 +7.7n - 7.8[/tex]
(ii) Given,
Diameter of the core = the first term of the AP = [tex]a_1[/tex]
Twice the thickness of the paper = common difference= d
When one sheet of paper is rolled over the core,
Then the new diameter of the roll = [tex]a_1+d[/tex], the second term of the AP([tex]a_2[/tex])
To find [tex]a_2[/tex]
We have [tex]S_n = 0.1n^2 + 7.9n[/tex]
[tex]S_1 = first term = a_1 = 0.1x1 +7.9 = 0.1+7.9 = 8[/tex]
[tex]S_2 = a_1 + a_2[/tex]
[tex]0.1 x 2^2 + 7.9 x2 = 8 +a_2[/tex]
[tex]0.4+15.8 = 8+a_2[/tex]
[tex]16.2 = 8+a_2[/tex]
[tex]a_2 = 8.2[/tex]
Hence, the diameter of the roll when one sheet is rolled over = 8.2cm