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geometric progession help

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    first two terms of geometric progression are 10 and 4
    i. find the sum of the first 15 terms
    ii. find the sum to infinity in exact form

    so sn=a(r^n-1)/r-1 a=10 n=15 and r=0.4

    s15=10(0.4^14)/0.4-1= -4.473924267*10^-2 is this right?
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    No it's not.

    You seem to be missing a -1 in the numerator and it's to the power 15.
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    it is r^n-1
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    (Original post by dongonaeatu)
    first two terms of geometric progression are 10 and 4
    i. find the sum of the first 15 terms
    ii. find the sum to infinity in exact form

    so sn=a(r^n-1)/r-1 a=10 n=15 and r=0.4

    s15=10(0.4^14)/0.4-1= -4.473924267*10^-2 is this right?
    No.

    \displaystyle s_n=a\frac{r^n-1}{r-1} \rightarrow s_{15}=10\cdot \frac{0.4^{15}-1}{0.4-1}
    and the sum of infinity for |r|<1

    \displaystyle s=a\frac{1}{1-r}
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    (Original post by dongonaeatu)
    first two terms of geometric progression are 10 and 4
    i. find the sum of the first 15 terms
    ii. find the sum to infinity in exact form

    so sn=a(r^n-1)/r-1 a=10 n=15 and r=0.4

    s15=10(0.4^14)/0.4-1= -4.473924267*10^-2 is this right?
    The formula for sum is,  \displaystyle S_n =\frac{a(r^{n}-1)}{r-1}

    So inputting the values gives,
     \displaystyle S_{15}=\frac{10(0.4^{15}-1)}{0.4-1}
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    (Original post by ztibor)
    No.

    \displaystyle s_n=a\frac{r^n-1}{r-1}
    and the sum of infinity for |r|<1

    \displaystyle s=a\frac{1}{1-r}
    i got 1.666638771 but i guess thats wrong too as its the sum of the first 15 terms and the first term is 10
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    (Original post by raheem94)
    The formula for sum is,  \displaystyle S_n =\frac{a(r^{n}-1)}{r-1}

    So inputting the values gives,
     \displaystyle S_{15}=\frac{10(0.4^{15}-1)}{0.4-1}
    raheem you tank

    is the answer 16.66664877
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    (Original post by ztibor)
    No.

    \displaystyle s_n=a\frac{r^n-1}{r-1} \rightarrow s_{15}=10\cdot \frac{0.4^{15}-1}{0.4-1}
    and the sum of infinity for |r|<1

    \displaystyle s=a\frac{1}{1-r}
    thats wrong the a should be above the division line
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    (Original post by dongonaeatu)
    raheem you tank

    is the answer 16.66664877
    Yes
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    (Original post by dongonaeatu)
    thats wrong the a should be above the division line
    It doesn't matters, ztibor is correct.
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    (Original post by raheem94)
    It doesn't matters, ztibor is correct.
    ah okay. What is the formula for the sum to infinity
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    (Original post by dongonaeatu)
    ah okay. What is the formula for the sum to infinity
    ztibor has given you that
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    (Original post by ztibor)
    No.

    \displaystyle s_n=a\frac{r^n-1}{r-1} \rightarrow s_{15}=10\cdot \frac{0.4^{15}-1}{0.4-1}
    and the sum of infinity for |r|<1

    \displaystyle s=a\frac{1}{1-r}
    it says find the sum to infinity in exact form so.

    sum to infinity= 10/1-0.4= 16.6?
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    (Original post by TenOfThem)
    ztibor has given you that
    is the sum to infinity 16.6
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    10/0.6 = 100/6 = 50/3

    you seem to have rounded so that would not be exact form
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    (Original post by raheem94)
    It doesn't matters, ztibor is correct.
    is the sum to infinity 16.6
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    (Original post by dongonaeatu)
    is the sum to infinity 16.6
    Yes, it is 16.7(to 3s.f.).
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    (Original post by TenOfThem)
    10/0.6 = 100/6 = 50/3

    you seem to have rounded so that would not be exact form
    yeah because it had lots of 6's in it lol. can i just leave it as 50/3 then?
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    (Original post by dongonaeatu)
    yeah because it had lots of 6's in it lol. can i just leave it as 50/3 then?
    The question asks for exact answer hence you MUST write it as  \displaystyle \frac{50}3 \ or \ 16\frac23
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    (Original post by raheem94)
    The question asks for exact answer hence you MUST write it as  \displaystyle \frac{50}3 \ or \ 16\frac23
    hey man, solve 3*5^x=150 giving it to 4dp

    is it; log3*5^x=log150

    5^x=log150/log3

    5^x=4.560876795 then i dont know how to get the x on its own
 
 
 
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