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Thread: geometric progression

  1. #1
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    geometric progression

    In a geometric progression the sixth term is 8 times the third term and the sum of the 17th and 8th term is 192.
    Determine (i) The common ratio (ii) The first term (iii) The sum of the 5th to 11th terms inclusive

    solution: Given that in the geometric series sixth term is 8 times the third term
    i.e. t6 =8t3 .. (1)
    And the sum of the 17th and 8th term is 192
    i.e. t 17 +t 8 =192 (2)

    (i)common ratio r=2
    (ii) first term a= 192/65664 =0.003
    (iii)
    Sn= a(1-r^n)/(1-r)
    number of terms between 5 to 11 is=7
    S7= 0.003(127)=0.381
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by gracy View Post
    In a geometric progression the sixth term is 8 times the third term and the sum of the 17th and 8th term is 192.
    Determine (i) The common ratio (ii) The first term (iii) The sum of the 5th to 11th terms inclusive

    solution: Given that in the geometric series sixth term is 8 times the third term
    i.e. t6 =8t3 ….. (1)
    And the sum of the 17th and 8th term is 192
    i.e. t 17 +t 8 =192 …… (2)

    (i)common ratio r=2
    (ii) first term a= 192/65664 =0.003
    You should include some explanation, but this checks out, though you should
    keep more significant digits in (ii), I make it 0.002924

    (iii)
    Sn= a(1-r^n)/(1-r)
    number of terms between 5 to 11 is=7
    S7= 0.003(127)=0.381
    Sn=a(1-r^n))/(1-r).

    The sum you seek is S11 - S4 = [(1-r^11) - (1-r^4)] a /(1-r)=5.94..
    RonL
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  3. #3
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    Hello, Gracy!

    Check the original problem.
    . . I'm certain there is a typo.
    My corrected version is much nicer problem . . .


    In a geometric progression the sixth term is 8 times the third term
    and the sum of the 7th and 8th term is 192.
    Determine: (i) The common ratio
    . . . . . . . . (ii) The first term
    . . . . . . . . (iii) The sum of the 5th to 11th terms inclusive

    The sixth term is: .a(6) = ar^5
    The third term is: .a(3) = ar^2

    We are told: .a(6) .= .8a(3) . . ar^5 .= .8ar^2 . . r = 8 . . r = 2 .(i)


    The 7th term is: .a(7) .= .ar^6
    The 8th term is: .a(8) .= .ar^7

    Their sum is: .ar^6 + ar^7 .= .ar^6(1 + r) .= .192

    Substitute (i): .a2^6(1 + 2) .= .192 . . a = 1 .(ii)


    Now that we know the sequence: .1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, . . .

    the sum of the 5th to 11th terms is:
    . . 16 + 32 + 64 + 128 + 256 + 512 + 1024 .= .2032 .(iii)

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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Hey guys,

    Just curious. Is there a formula for finding the sum of a given range of terms? For instance, if the question asked: find the sum of the 5th to the 78th term inclusive, how would you do that? Or is that a type of question that will never be asked?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    Hey guys,

    Just curious. Is there a formula for finding the sum of a given range of terms? For instance, if the question asked: find the sum of the 5th to the 78th term inclusive, how would you do that? Or is that a type of question that will never be asked?
    The sum of the first n terms is Sn=a (1-r^n)/(1-r), so the sum S(m,n) of the
    n to m'th term is the sum of the first m terms minus the sum of the first n-1
    terms so:

    S(m,n) = a/(1-r) [1-r^m - 1 + r^(n-1)] = a/(1-r) [r^(n-1)-r^m]

    RonL
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    Hey guys,

    Just curious. Is there a formula for finding the sum of a given range of terms? For instance, if the question asked: find the sum of the 5th to the 78th term inclusive, how would you do that? Or is that a type of question that will never be asked?
    5th term = a*k^4

    6th term = a*k^5

    ...

    78th term = a*k^77

    Thus,

    Sum = a*k^4 + a*k^5+...+a*k^77
    Factor,
    a*k^4(1+k+k^2+...+k^77)
    This is geometric series again.
    You can use formula if k not = to 1.
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