急……一道高中数学题：数列（要过程）

设a(1)、a(2)、a(3)、a(4)、a(5)……、a(n)，公差是a(1)-a(2)。
3a(4)=7a(7)
3a(4)=7a(4)-3[a(1)-a(2)]
0=4a(4)-3a(1)+3a(2)
4a(4)+3a(2)=3a(1)
4{a(1)-3[a(1)-a(2)]}+3a(2)=3a(1)
4a(1)-12a(1)+12a(2)+3a(2)=3a(1)
15a(2)-11a(1)=0
15a(2)=11a(1)
a(2)=11a(1)/15
Sn=a(1)+a(2)+a(3)+a(4)+a(5)……+a(n)
=[a(1)+a(n)]×a(n)÷2
={a(1)+(n-1)[a(1)-a(2)]}×(n-1)[a(1)-a(2)]÷2
=[a(1)+a(1)(n-1)-a(2)(n-1)]×[a(1)(n-1)-a(2)(n-1)]÷2
=[a(1)+na(1)-a(1)-na(2)+a(2)]×[na(1)-a(1)-na(2)+a(2)]÷2
=[na(1)-na(2)+a(2)]×[na(1)-a(1)-na(2)+a(2)]÷2
={n[a(1)-a(2)]+11a(1)/15}×{n[a(1)-a(2)]-a(1)+11a(1)/15}÷2
=({n[a(1)-a(2)]^2-a(1)n[a(1)-a(2)]+11a(1)/15 n[a(1)-a(2)]/15}+11a(1)/15×[na(1)-na(2)]-11a(1)^2/15+121a(1)^2/225)÷2
={-2a(1)a(2)n^2+na(2)^2+4na(1)a(2)/15+132na(1)^2/225-11na(1)a(2)/225+11na(1)^2/15-11a(1)^2/15}÷2
=(66n/225)[a(1)^2]+(49/450-n)[na(1)a(2)]+[na(2)^2]/2
n肯定越大越好。