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anonymous

  • one year ago

The first three terms of an arithmetic sequence, a, a+d and a + 2d, are the same as the first three terms, a, ar, ar^2, of a geometric sequence ( a does not equal 0). Show that this is only possible if r=1 and d=0 I am not sure how they want me to prove this.

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  1. Koikkara
    • one year ago
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    @imqwerty this user might help you. @Zas

  2. anonymous
    • one year ago
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    Thanks!

  3. imqwerty
    • one year ago
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    the sum of 1st 3 terms of AP1st term of both the series is a so its clear that a=a :D now we come to the second term\[a+d=ar\] \[d=ar-a=>d=a(r-1)\] nd the third terms are also equal so \[a+2d=ar^2\] put d=a(r-1) in this eq. \[a+2a(r-1)=ar^2\]divide the whole equation by a u get-\[1+2(r-1)=r^2 =>r^2 -2r+1=0 =>(r-1)^2=0\]so r =1 put r=1 in d=a(r-1) d=a(1-1) d=0

  4. imqwerty
    • one year ago
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    hey forget that 1st line i forgot to delete that instead of that it must be- 1st terms of both series r same so a=a

  5. anonymous
    • one year ago
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    Ok, thank you very much!

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