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angela210793

  • 5 years ago

I'll check the answer tomorrow after school :) Help please :) There's a geometric progression....a7-a3=60 and a7-a5=48 what is S5? q>0

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  1. watchmath
    • 5 years ago
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    \(a_7=a_1r^6\) and \(a_3=a_1r^2\) So \(a_1r^6-a_1r^2=a_1r^2(r^4-1)=60\) Using similar argument \(a_1r^6-a_1r^4=a_1r^4(r^2-1)=48\) It follows that \(60/48=\frac{a_1r^2(r^4-1)}{a_1r^4(r^2-1)}=\frac{r^2+1}{r^2}\) Cross multiply we have \(60r^2=48r^2+48\) \(12r^2=48\) \(r^2=4\) \(r=\pm 2\) What is the question again? S5? the sum of the first 5 terms?

  2. angela210793
    • 5 years ago
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    yea tht's the question

  3. angela210793
    • 5 years ago
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    wht is r? is it y2/y1=r? cause we use q..... :)

  4. watchmath
    • 5 years ago
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    oh r is the ratio

  5. watchmath
    • 5 years ago
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    Ok so plug in \(r=\pm 2\) to \(a_1r^2(r^4-1)=60\) we have \(a_1\cdot 4\cdot (16-1)=60a_1=60\) so \(a_1=1\) Now knowing \(a_1=1\) and \(r=\pm 2\) you can find your S5.

  6. angela210793
    • 5 years ago
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    thanks ^_^ :)

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