S1=1+3=4=a1
2a1+d(n–1)
Sn = ---------------- • n
2
8+d(n–1)
--------------- • n = n^2 + 3n |•2
2
n•(8+dn–d) = 2•(n^2+3n)
8n+dn^2–dn = 2n^2 + 6n
2n^2–dn^2+dn–2n = 0
2n(n–1)–dn(n–1) = 0
(n–1)(2n–dn) = 0
n–1=0; 2n–dn = 0
n=1; n(2–d) = 0
n=0; 2–d=0
d=2
Ответ: d=2