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anonymous

  • one year ago

I need some help with the last question of an exercise thats too long to type out so I will attach a picture. The answers for all the questions before are: (i) = a+4d; a+14d (iii) This is the one I need help with and I know that the answer is 2.4 but I can't get there.

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  1. anonymous
    • one year ago
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    I am sorry I made a mistake the answer is 2.5. still need help though

  2. ganeshie8
    • one year ago
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    Have you finished part ii ?

  3. anonymous
    • one year ago
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    Yes

  4. ganeshie8
    • one year ago
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    good, then simply take the ratio of second term and first term : \[\dfrac{a+4d}{a}\] plugin \(d = \dfrac{3}{8}a\) and simplify

  5. anonymous
    • one year ago
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    Ah OK thanks.

  6. ganeshie8
    • one year ago
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    good, then simply take the ratio of second term and first term : \[\text{common ratio}=\dfrac{a+4d}{a}\] plugin \(d = \dfrac{3}{8}a\) and simplify

  7. ganeshie8
    • one year ago
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    notice that \(a\) cancels out

  8. anonymous
    • one year ago
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    so I get 1.5a ?

  9. ganeshie8
    • one year ago
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    plugging \(d = \dfrac{3}{8}a\) gives \[\text{common ratio}=\dfrac{a+4d}{a} = \dfrac{a+4*\dfrac{3}{8}a}{a}=1+\dfrac{3}{2}=?\]

  10. anonymous
    • one year ago
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    Oh ok! So the a's cross out to 1 which is added?

  11. ganeshie8
    • one year ago
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    Yes, we can factor out \(a\) and cancel it with the \(a\) in the bottom : \[\text{common ratio}=\dfrac{a+4d}{a} = \dfrac{a+4*\dfrac{3}{8}a}{a}=\dfrac{a\left(1+4*\dfrac{3}{8}\right) }{a}=1+\dfrac{3}{2}=?\]

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