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
A)15cosx=3cosx*5sinx
15cosx/cosx=3cosx/cosx*5sinx/cosx
15=3*5tgx
5tgx=5
tgx=1
x=П/4+Пn
б)5П<П/4+Пn<13П/2
5П-П/4<П/4+Пn-П/4<13P/2-П/4
10P/4-P/4<Pn<26P/4-P/4
9P/4<Pn<25P/4
2,25<n<6,25
n=3,4,5,6
Объяснение:
Решение смотри в фотографиях