- 21 May '04 15:48i 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...) - 23 May '04 10:31 / 1 edit

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...*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...)

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 - 25 May '04 23:13

Let x(n) be a sequence.*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

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. - 25 May '04 23:36

sum A to n terms: na + n(n-1)*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 B to n terms: a(2^(n+1)-1)

Equal at n=4

a(2^5-1) = 4a + 12

=> a = 12/27 = 4/9

RTF n such that

2na + 2n(n-1) < a(2^(n+1)-1)

2n(1 + 9/4*(n-1)) < 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 > 127

n=7: LHS = 14*(1+27/2) = 7*29 = 203 < 255

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