Hard arithmetic progression problem

[Guest]

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 I am not getting the desired result.
Could someone please help me?

[HallsofIvy]

We are told this is an arithmetic sequence so the first n terms are
a, a+ r, a+ 2r, ..., a+ (n-2)r, a+ (n-1)r, a+ nr

To sum those, use a trick that is attributed to Gauss (as a child!). Reverse them
a+ nr, a+ (n-1)r, a+ (n-2)r, ..., a+ 2r, a+ r, a and add vertically to get
a+ a+ nr= 2a+ nr, a+r+a+ (n-1)r= 2a+ nr, a+ 2r+ a+ (n-2)r= 2a+ nr....
Get the point? all those vertical sums give 2a+ nr so the sum of both rows is (2a+ nr)n= 2an+ n^2r. The two rows are the same, of course so the sum of just one is an+ n^2r/2.

We can do the same thing with the last n terms:
a+ (n+1)r, a+ (n+2)r..., a+ (2n-1)r, a+ 2nr and reverse
a+ 2nr, a+ (2n- 1)r, ..., a+ (n+2)r, a+ (n+ 1)r. Summing vertically,
a+ (n+ 1)r+ a+ 2nr= 2a+ (3n+ 1)r, a+ (n+2)r+ a+ (2n-1)= 2a+ (3n+1)r, etc.
Now all n "vertical" sums add to 2a+ (3n+ 1)r so the two rows sum to (2a+ (3n+ 1)r)n= 2an+ 3rn^2+ nr. So each row adds to an+ (3/2)rn^2+ rn/2.

The condition is that the sum of the last n terms is three times the sum of the first n terms. That is
an+ (3/2)rn^2+ rn/2= 3(an+ n^2r/2).