Understanding Paper Roll Length Increase
When a long ream of paper with thickness $t$ is rolled tightly, each turn adds a layer of thickness $t$ to the radius of the roll.
Calculating Circumference Change
Let the radius of the roll be $r$ before adding a new turn. The length of the paper in this turn is approximately equal to the circumference, $C_1 = 2\pi r$.
After adding one full turn of paper (thickness $t$), the new radius becomes $r + t$. The length of the paper in this new, outer turn is approximately the new circumference, $C_2 = 2\pi (r + t)$.
Determining the Difference in Length
The difference in length between the current turn and the previous turn is:
$ \Delta L = C_2 - C_1 $
$ \Delta L = 2\pi (r + t) - 2\pi r $
$ \Delta L = 2\pi r + 2\pi t - 2\pi r $
$ \Delta L = 2\pi t $
Therefore, the length of the paper wrapped in one turn exceeds the length in the previous turn by $2\pi t$.
Important Questions from Progression (Notes)
- The sum of 16 terms of the series $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + .....$ is :
- An auditorium has 8 seats in the first row, with every row to follow having 4 more seats than its preceding row. The total capacity is 416. What is the minimum number of rows needed to seat 150 people?
- The $5^{\text{th}}$ and $9^{\text{th}}$ terms of an arithmetic progression are 7 and 13 respectively. What is the $15^{\text{th}}$ term?
- Find the sum of the G.P.:
$5/11, 5/121, 5/1331, 5/14641, ...$ to $n$ terms. Suppose $a_1, a_2,..., a_{300}$ are integers such that $a_{i-1}+ a_i+ a_{i+1} = 2025$ for all $i = 2,3, ..., 299$.
If $a_7 = -5, a_9 = 37$, then the value of $a_{106}$ is
