- #1
Purpleshinyrock
- 27
- 6
Summary:: Sequences, Progressions
Hello. I have been Given the following exercise, Let (a1, a2, ... an, ..., a2n) be an arithmetic progression such that the sum of the last n terms is equal to three times the sum of the first n terms. Determine the sum of the first 10 terms as a function of the ratio d.
Solution is S10=50d
I know that an=a1+nd-d, and an+1=a1+nd
a2n=a1+(2n-1)d=a1+2nd-d
sn=(a1+an)(n/2)=(2a1+nd-d)(n/2)
Sn=(an+1+a2n)(n/2)=(2a1+3nd-r)(n/2)
But When I try to do Sn=3sn(the sum of the last n terms is equal to three times the sum of the first n terms) I am not getting the desired result.
Could someone please help me?
[Moderator's note: Moved from a technical forum and thus no template.]
Hello. I have been Given the following exercise, Let (a1, a2, ... an, ..., a2n) be an arithmetic progression such that the sum of the last n terms is equal to three times the sum of the first n terms. Determine the sum of the first 10 terms as a function of the ratio d.
Solution is S10=50d
I know that an=a1+nd-d, and an+1=a1+nd
a2n=a1+(2n-1)d=a1+2nd-d
sn=(a1+an)(n/2)=(2a1+nd-d)(n/2)
Sn=(an+1+a2n)(n/2)=(2a1+3nd-r)(n/2)
But When I try to do Sn=3sn(the sum of the last n terms is equal to three times the sum of the first n terms) I am not getting the desired result.
Could someone please help me?
[Moderator's note: Moved from a technical forum and thus no template.]
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