[RESOLVED] Arithmetic and Geometric Sequences Problems

Ahh it's been awhile after I got to know how to solve differentiation problems.. Anyways, I bumped into a few tough questions of arithmetic and geometric sequences.

1. An arithmetic progression has first term a and common difference -1. The sum of the first n terms is equal to the sum of the first 3n terms. Express a in terms of n. (Answer: 2n - 1/2)

2. The sum of the first hundred terms of an arithmetic progression with the first term a and common difference d is T. The sum of the first 50 odd-numbered terms,i.e. the first, third, fifth,.....,ninety-ninth, is 1/2T - 1000. Find the value of d. (Answer: 40)

3. A ball is dropped from a height of 1 metre onto a level table. It always rises to a height equal to 0.9 of the height from which it was dropped. How far does it travel in total until it stops bouncing? (Answer: 19m)

From all the questions on arithmetic and geometric progression, I was stuck at these 3 questions... Hope someone can help me out... :ehh:

1. Sum to n terms = (n/2)(2a + (n - 1)d)

d = -1

= (n/2)(2a - (n - 1))

Sum to 3n terms = (3n/2)(2a - (3n - 1))

(3n/2)(2a - (3n - 1))= (n/2)(2a - (n - 1))

3(2a - 3n + 1)=(2a - n + 1)

Rearrange to get answer

2.

T = (100/2)(2a + (100 - 1)d) = 100a + 4950d from first line (1)

Odd number terms have same first term but the difference is doubled and have 50 terms so:

1/2T - 1000 = (50/2)(2a + (50 - 1)2d) = 50a + 1225d

Double to get

T - 2000 = 100a + 2450d (2)

Eliminate T and a from (1) and (2)

3. Geometric series a = 1, r = 0.9

Same question as finding the sum to infinity of the series

Yay once again thank you! I've learned a lot :D

Oh btw, for question 2, the 1/2T - 1000 = 50a + 2450d

Hee.. maths I can do simple calculations but when it comes to application part, I simply have no idea what to do... :confused:

Yeah you're right sorry. I missed the 2 out of 2d

Interesting questions again though I've seen 3 before

3) This is how I would do it.

First it starts off at 1m, then drops, then each time after that, its 0.9 of the previous height. Also, the total distance travelled is 2* the current height, since it goes up, then back down. So it would be something like this:

1+2* (sum of 0.9^i), and do the sum until 0.9^i is really small.

There is something I learned in first year calc about the convergence of series. I'll try and dig it up for you and post it.

[EDIT]
Ok, I found the equation I was talking about. Since this is a geometric series, it can be expressed as sum(a*r^n). In this case, a=1, r=0.9. The series converges is |r|<1, which is is, so the sum is:

x/(1-r) where x is the first term, r as above.

So in this case, x=0.9, so the FULL sum of the ball bouncing is:

1 + 2 * (0.9 / (1 - 0.9) )
1 + 2 * (0.9 / 0.1)
1 + 2 * (9)
=19m

Which is finding the sum to infinity of the series as I said

O.o so complicated lol.... Give me some time to understand... :blush:

[EDIT] Yay finally managed to understand everything!! Thank you guys :bigyello: