Results 1 to 9 of 9

Math Help - Arithmetic Progression & Geometric Progression

  1. #1
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108

    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
    Attached Thumbnails Attached Thumbnails Arithmetic Progression & Geometric Progression-img_0001.jpg  
    Last edited by maybeline9216; March 19th 2009 at 04:00 AM. Reason: .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by maybeline9216 View Post
    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
    Please show your working, and the place you stuck in
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108
    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..
    Attached Thumbnails Attached Thumbnails Arithmetic Progression & Geometric Progression-img_0002.jpg  
    Last edited by maybeline9216; March 19th 2009 at 05:23 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2009
    Posts
    3
    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..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    3
    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..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by aamberlim View Post
    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

    Quote Originally Posted by maybeline9216 View Post

    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member maybeline9216's Avatar
    Joined
    Sep 2008
    From
    Singapore~
    Posts
    108
    Quote Originally Posted by aamberlim View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Like a stone-audioslave ADARSH's Avatar
    Joined
    Aug 2008
    From
    India
    Posts
    726
    Thanks
    2
    Quote Originally Posted by maybeline9216 View Post

    but Sn-3 simply means the sum between the 3rd term and the last term which is n term...ehh..im so confused lol.
     <br />
S_{n} = a_1 +a_2 \cdots + a_{n} <br />

     <br />
S_{n-3} = a_1 +a_2 \cdots + a_{n-3} <br />

    Substract the two equations

    Game Over XX!!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Mar 2009
    Posts
    3
    Quote Originally Posted by maybeline9216 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 24th 2010, 09:20 AM
  2. Arithmetic Progression or Arithmetic Series Problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 8th 2009, 01:36 AM
  3. Arithmetic & Geometric Progression
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 20th 2009, 09:00 PM
  4. Arithmetic & Geometric Progression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 5th 2008, 08:16 AM
  5. Arithmetic Progression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 14th 2008, 10:11 AM

Search Tags


/mathhelpforum @mathhelpforum