Sequence and Series (1 Viewer)

[velox]

im a bit lost, easy question but i cant get it.....
Find the 7th term of the arithmetic progression whose 5th term is m and whose 11th term is n.


Also i cant do this one
How many terms are in the sequence 9, 12, 15,...,(6p+15)

thanks

 

[Heinz]

For the first one
t₅ = a + 4d = m
t₁₁ = a + 10d = n
n-m = 6d
d= [n-m]/6
m+n = 2a + 14[n-m]/6
6m + 6n = 12a + 14n -14m
20m - 8n = 12a
5m - 2n = 3a
a = [5m - 2n]/3

t₇ = a + 6d
= [5m - 2n]/3 + n-m
= [5m - 2n + 3n -3m]/3
= [2m + n]/3

For the second one
note that its an AP
t₁ = a = 9
t₂ = a + d = 12
:. d = 3

t_n = a + (n-1)d
= 9 + 3n - 3
= 6 + 3n
6p + 15 = 6 + 3n
3n = 9 + 6p
n = 3 + 2p

Hope these are right

 

[velox]

i dont get how u got this 6p + 15 = 6 + 3n, maybe cos its late at nite i dunno...sumone plz explain!!

 

[~dEjA vOuX~]

6p + 15 = 9 + (n-1).3
6p + 15 = 9 + 3n - 3
6p + 15 = 6 + 3n
...

Does it makes sense now?

 

[velox]

15. Find the first value of n for which the series 227 + 225 + 223 + becomes negative.

My Working:

tn = (a + (n 1) d) < 0
tn = (227+ (n 1)(-2)) < 0

227 2n + 2 < 0
229 - 2n < 0
229 < 2n
114.5 <n
n>114.5
n=115

t115 = 227 +114(-2)
= 227 228
= -1

What have I done wrong? The answer is 115. I see that I have gone to far in working out, but I did the same for a very similar question and got the right answer, as shown below.

14. Find the first value of n for which the series 100 + 93 + 86 + becomes negative.

My Working:

tn = (a + (n 1)d) < 0
tn = (100 + (n 1)(7)) < 0

100 + 7n 7 <0
93 + 7n < 0
7n < -93
n<13 2/7
n=13 gives us the first negative term

t18 = 100 + (12) (-7)
t18 = 100 84
t18 = 16

Here the answer is 16, which is what I worked it out to be.

Btw thanks for the above posts

 

[Heinz]

Originally posted by wrx
i dont get how u got this 6p + 15 = 6 + 3n, maybe cos its late at nite i dunno...sumone plz explain!!

Your given that the final term is 6p + 15 and you know that the final term is also a + (n-1)d. Sub in the values for a, n and d then equate them.


for 15. Theres nothing wrong with your working for this question. You found the value of n and substituting that value serves as a means of verification. Thats fine, shows your right.

For 14, you made a small error in the second line. The difference is minus 7.

t_n = a + (n 1)d
100 + (n 1)(-7) < 0

100 - 7n + 7 < 0
107 < 7n
n > 107/7
n = 16 the first integer after 107/7

 

[velox]

ah ok kool thanks, ah ic now, question 15 was asking for first negative value, and the verification bit proved that with the -1 as the term. Now i understand

 

[velox]

btw is sequence in the syllabus?

Actually just found out from teacher, it isnt, but as CM_Tutor said, they are interchangeable in the book so it would be best to do it anyway

 

[Estel]

Yes. Topic 7.

 

[CM_Tutor]

Note that most HSC texts use the terms 'sequence' and 'series' as if they are interchangeable (which they aren't), so I wouldn't get too concerned about the difference.

 

[liddy]

a logical method?

heya
i was just wondering whether when asked to write the expression for the nth term of the series, is there a method to work this out or is it just trial and error/looking at it?

eg. - 21 - 17 - 13 - ...
I know the answer is Tn = 4n - 25, but i dont actually know how to logically work it out.

anyone know?
ta!

 

[CM_Tutor]

T_n = a + (n - 1)d where a is the first term and d is the common difference.

In this case, a = -21, d = 4. So, T_n = -21 + (n - 1)(4) = -21 + 4n - 4 = 4n - 25