# Thread: Arithmetic Progression & Geometric Progression

1. ## 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 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

2. 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 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

3. 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..

4. 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 (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..

5. 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..

6. 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 (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..
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

Originally Posted by maybeline9216

Note:For better view, pls double click and zoom in.thks
I am really sorry 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

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

7. 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.

8. 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!!

9. 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?

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# d first nd last term of an AP are 1 nd 121 respectively find d no of term in d Ap

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