Question

3.Найдите число членов арифметической прогрессии, если a3-a1=8; a2+a4=14.Sn=111​

Answer 1

a(n) = a(1) + d(n - 1)

a(n) = a(n - 1) + d

====

a(3) - a(1) = 8

a(2) + a(4) = 14

S(n) = 111

n - ?

a(1) + 2d - a(1) = 8

a(1) + d + a(1) + 3d = 14

2d = 8

2a(1) + 4d = 14

d = 4

a(1) = -1

S(n) = (2a(1) + d(n - 1))/2 *n

222 = (-2 + 4(n - 1))*n

222 = (-2 + 4n - 4))*n

111 = (2n - 3)n

2n^2 - 3n - 111 = 0

D = 9 + 4*2*111 = 897

√897 - решений нетпри таких условиях

a(n) = -1, 3, 7, 11, 15, 19, 23, 27, 31

S(8) = 104

S(9) = 135