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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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.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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@imqwerty this user might help you. @Zas
Thanks!
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

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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
Ok, thank you very much!

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