# Arithmetic Progression & Geometric Progression

• Mar 19th 2009, 03:58 AM
maybeline9216
Arithmetic Progression & Geometric Progression
Please teach me how to do qn 8 and 9. Qns 8 and 9 are attached.

And also this qn as well:D Thks alot!!

Qn 3) The sum of the first 3 terms and that of the last 3 terms of an AP are 9 and 57 respectively.It is known that the sum of the whole sequence is 121.Find the number of terms in this sequence.

Note:For better view, pls double click and zoom in.thks
• Mar 19th 2009, 04:58 AM
Quote:

Originally Posted by maybeline9216
Please teach me how to do qn 8 and 9. Qns 8 and 9 are attached.

And also this qn as well:D Thks alot!!

Qn 3) The sum of the first 3 terms and that of the last 3 terms of an AP are 9 and 57 respectively.It is known that the sum of the whole sequence is 121.Find the number of terms in this sequence.

Note:For better view, pls double click and zoom in.thks

• Mar 19th 2009, 05:11 AM
maybeline9216
Here r my workings for qn 3...but im stuck..

i know how to do qn 8 already..pls teach me qn 9..i have no idea how to approach qn 9..
• Mar 21st 2009, 09:34 PM
aamberlim
For question 3,

First 3 terms,
a + d = 3 ---- (1)

Last 3 terms,
Sn - Sn-3 = 57

Since Sn = 121,
121 - Sn-3 = 57
Sn-3 = 64

Sn = n/2 {2a + (n-1)d}
242 = n{2a +dn-d}
2an + dnē- dn = 242 ----(2)

sub (1) into (2),

d= (242-6n)/(nē-3n) ----(3)

Sn-3 = (n-3)/2 {2a + (n-4)d} = 64 ----(4)

sub (3) into (4),

132n = 1452
Therefore, n = 11 (Clapping) (yay! hahas..)

er.. is the working complicated?

btw, how you do qn 7(ii)? cos i didn't to figure out how to start the 2% thing.. (Shake)
• Mar 21st 2009, 10:52 PM
aamberlim
hahas.. paiseh.. manage to do qn 7(ii) le..

For question 9,

i) common ratio, r = 0.9
Tn = ar^n-1

interval between 1st and 2nd bounce -> 2s
therefore, a = 2
T1 = 2(0.9)^0 = 2

interval between nth and 9n+1)th bounce -> ?
Tn = ar^n-1
0.02 < 2(0.9)^n-1
0.01 < 0.9^n-1
0.01 < (0.9^n)/0.9
lg 0.009 < lg 0.9^n
n > 44.71 (to 4s.f.)
therefore, n = 45

ii) Sn = (a(1-r^n)) / (1-r) = (2(1-0.9^45)) / (1-0.9) = 19.8 s (to 3s.f.)

sorry if the workings came late.. cos i'm also rushing through the whole tutorial to hand up to my tutor on mon during maths tutorial..
• Mar 21st 2009, 11:02 PM
Quote:

Originally Posted by aamberlim
For question 3,

First 3 terms,
a + d = 3 ---- (1)

Last 3 terms,
Sn - Sn-3 = 57

Since Sn = 121,
121 - Sn-3 = 57
Sn-3 = 64

Sn = n/2 {2a + (n-1)d}
242 = n{2a +dn-d}
2an + dnē- dn = 242 ----(2)

sub (1) into (2),

d= (242-6n)/(nē-3n) ----(3)

Sn-3 = (n-3)/2 {2a + (n-4)d} = 64 ----(4)

sub (3) into (4),

132n = 1452
Therefore, n = 11 (Clapping) (yay! hahas..)

er.. is the working complicated?

btw, how you do qn 7(ii)? cos i didn't to figure out how to start the 2% thing.. (Shake)

From the infinite sum formula get the value of common ratio
Put the values in the equation for n terms
And then try solving the inequality

Quote:

Originally Posted by maybeline9216

Note:For better view, pls double click and zoom in.thks

I am really sorry :o for late reply but I was uncomfortable with the attatchments ,I had to save and zoom them to a lesser extent , I think you could type such questions they aren't big ones (Nod)

For the question 9

Time for bounces are

2, 2*9/10, 2*9^2/10^2, 2*9^3/10^3......

Solve this inequality

$2\times (\frac{9}{10})^{n-1} \le 0.002$

Notice that LHS is nothing other than the nth term of GP

ii)
The sum of nth terms of GP will be used
• Mar 22nd 2009, 07:11 AM
maybeline9216
Quote:

Originally Posted by aamberlim
For question 3,

Last 3 terms,
Sn - Sn-3 = 57

Hey..this doesn't make sense to me...Sn means the total sum of n terms rite...then S3 is the sum of the first 3 terms...but Sn-3 simply means the sum between the 3rd term and the last term which is n term...ehh..im so confused lol.
• Mar 22nd 2009, 07:20 AM
Quote:

Originally Posted by maybeline9216

but Sn-3 simply means the sum between the 3rd term and the last term which is n term...ehh..im so confused lol.

$
S_{n} = a_1 +a_2 \cdots + a_{n}
$

$
S_{n-3} = a_1 +a_2 \cdots + a_{n-3}
$

Substract the two equations

Game Over XX!!
• Mar 23rd 2009, 08:26 AM
aamberlim
Quote:

Originally Posted by maybeline9216
Hey..this doesn't make sense to me...Sn means the total sum of n terms rite...then S3 is the sum of the first 3 terms...but Sn-3 simply means the sum between the 3rd term and the last term which is n term...ehh..im so confused lol.

n is a number which, in this case, indicates the last term. so n-3 is another number, not the sum between the 3rd term and the last term.

The terms:

T1, T2, T3, T4, ..........,Tn-3, Tn-2, Tn-1, Tn

since the question say that the sum of last three terms is 57,

Total sum - Sum of 1st term to the last 4th term = 57
Therefore, Sn - Sn-3 =57

er.. can understand better? (Wondering)