 Posers and Puzzles

1. 21 May '04 15:48
i had my advanced higher maths paper today. it's a kind of combination of parts of the 1st and 2nd year uni courses. my head hurts...

i had 2 problems...one i kinda bluffed (i'll start anohter thread about thataone...), and then htere was this one. i managed all of it, apart form the little 2 mark bit at the end. twas surprisingly easy for the last question of the paper...

16. a. Obtain the sum of the series 8+11+14+...+56. (2)

b. A geometric sequence of positive terms has first term 2, and the sum of the first three terms is 266. Calcualte the common ratio. (3)

c. An arithmetic sequence, A, has first term a and common difference 2, and a geometric sequence, B, has first term a and common ration 2. The first four terms of each sequence have the same sum. Obtain the value of a. (3)

and now-the bit i couldn't do...for 2 marks...:

Obtain the smallest value of n such that the sum to n terms for sequence B is more than twice the sum to n terms for sequence A. (2)

there you go-so, can snyone solve the last bit? (the rest is easy, btw...)
2. 23 May '04 10:311 edit
Originally posted by genius
i had my advanced higher maths paper today. it's a kind of combination of parts of the 1st and 2nd year uni courses. my head hurts...

i had 2 problems...one i kinda bluffed (i'll start anohter thread about thataone...), and then htere w ...[text shortened]... u go-so, can snyone solve the last bit? (the rest is easy, btw...)
now if you would jst explain common difference and common ratio, then i think i'd understand better what the given means. I can't find a translation that makes sense in dutch...

EDIT

a) 8 + 11 + 14 +...+56 =
2*3 + 2 + 3*3 + 2 + 4*3 + 2 + ...+ 18*3 +2 =
(2+3+4+5+...+18)*3 + 17*2 =
(1/2 * 18 * 19 -1)*3 +34 =
170*3 + 34 = 544
3. 25 May '04 23:13
Originally posted by TheMaster37
now if you would jst explain common difference and common ratio, then i think i'd understand better what the given means. I can't find a translation that makes sense in dutch...

EDIT

a) 8 + 11 + 14 +...+56 =
2*3 + 2 + 3*3 + 2 + 4*3 + 2 + ...+ 18*3 +2 =
(2+3+4+5+...+18)*3 + 17*2 =
(1/2 * 18 * 19 -1)*3 +34 =
170*3 + 34 = 544
Let x(n) be a sequence.

x(n) is arithmetic if for some d x(n+1) - x(n) = d for all n. d is then called the common difference of the sequence.

x(n) is geometric if for some r x(n+1)/x(n) = r for all n. r is then called the common ratio of the sequence.
4. 25 May '04 23:36
Originally posted by genius
Obtain the smallest value of n such that the sum to n terms for sequence B is more than twice the sum to n terms for sequence A. (2)
sum A to n terms: na + n(n-1)
sum B to n terms: a(2^(n+1)-1)

Equal at n=4
a(2^5-1) = 4a + 12
=&gt; a = 12/27 = 4/9

RTF n such that
2na + 2n(n-1) &lt; a(2^(n+1)-1)
2n(1 + 9/4*(n-1)) &lt; 2^(n+1) -1
As it's only 2 marks, I assume you use trial and error at this point.
n=6: LHS = 12*(1+45/4) = 12 + 3*45 = 147 &gt; 127
n=7: LHS = 14*(1+27/2) = 7*29 = 203 &lt; 255

So the answer is n=7, if I've got this right. Probably not worth the effort for 2 marks though.