We can find A.P. in many situations in our day-to-day life. One such example is a tissue paper roll, in which the first term is the diameter of the core of the roll and twice the thickness of the paper is the common difference. If the sum of first n rolls of tissue on a roll is Sn = 0.1n2 +7.9n, then answer the following questions. (i) Find Sn – 1. (ii) What is the diameter of roll when one tissue sheet is rolled over it?

Answers 2

Answer:

(I) 0.1n2+7.7n-7.8

(2) 8.2cm

Step-by-step explanation:

hope this will help you!

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