Chapter 1 Text work review

10. Determine S₂₅.

     t₂ = 40, t₅ = 121     d = ?     t₁ = ?    t₂₅ = ?

      t₅ = t₂ + 3d  --> 121 = 40 + 3d --> 3d = 81 ---> d = 27

     t₂ = t₁ + d ---> t₁ = 13

     t₂₅ = 13 + 27(24) = 13 + 648 --> t₂₅ = 661

       S₂₅ = \( \frac{25}{2} \)(13 +  661) = \( \frac{25}{2} \)(674) = 8425

11. Find the first four terms:

   S₅ = 85   S₆ = 123   t₆ = ?   d = ?

    t₆ = S₆ -  S₅ = 123 - 85 = 38

   S₆ = \( \frac{6}{2} \)(t₁ + 38) = 123 ---> t₁ + 38 = 41 ---> t₁ = 3

    t₆ = 3 + 5d = 38 --> 5d = 35 --> d = 7

    so, the first four terms are 3, 10, 17, 24

12. Derive d = 5n².

time (s)	1	2	3	4
distance fallen in this second (m):	5	15	25	35
Total distance fallen:	5	20	45	80

We know that t₁ = 5, and t₂ = S₂ - S₁ = 20 - 5 = 15 = t₂

This gives us d = t₂ - t₁= 15 - 5 = 10 = d

and then, tn = t₁ + d(n - 1) = 5 + 10(n - 1) = 5 + 10n - 10 = 10n - 5 = t_n

Using S_n = \( \frac{n}{2} \)(t₁ + t_n) we get

          S_n = \( \frac{n}{2} \)(5 + 10n - 5) = \( \frac{n}{2} \)(10n) or

          S_n = 5n² = d

16. Distance from top to bottom...

First, let's be sure how many rings there are, according to diameter:

20, 19, 18, ..., 3

here, t₁ = 20, d = -1, and t_n = 3

3 = 20 - (n - 1) ---> n = 18, so there are 18 rings.

From the diagram, one can see that after the first ring, the distance from the bottom of one ring to the bottom of the next follows the pattern: 17, 16, 15, ..., 1, where there are 17 terms, with t₁ = 17, and t_n = 1, n = 17.

We want 20 + S₁₇ = 20 + \( \frac{17}{2} \)(17 + 1) = 20 + 153 = 173cm = distance from top to bottom.