- #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?

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