Геометрическая прогрессия

b2+b4=135, b5+b7=40 b1=? q=? S5=?

арифметическая прогрессия

a3+a4=7, a2+a8=13 a1 = ? d=? S10=?

B (n) = b1*q^ (n-1)

b2=b1*q

b4=b1*q^3

b5=b1*q^4

b7=b1*q^6

b1*q+b1*q^3=b1*q * (1+q^2) = 135

1+q^2=135 / (b1*q)

b1*q^4+b1*q^6=b1*q^4 * (1+q^2) = 40 подставляем

(b1*q^4*135) / b1*q=40 сокращаем

q^3=40/135 = (/5) = 8/27 q=2/3

b1 * (2/3) + b1 * (8/27) = 135

b1 * (2/3+8/27) = 135

b1 * (26/27) = 135 b1=3645/26

S (5) = (b1 * ((q^5) - 1) / (2/3-1) = (3645/26) * (-211/3^5) * (-3) = 9495/26

a (n) = a1+d * (n-1)

a2=a1+d

a3=a1+2*d

a4=a1+3*d

a8=a1+7*d

a1+2*d+a1+3*d=2*a1+5*d=7 2*a1=7-5*d

a1+d+a1+7d=2*a1+8*d=13 подставляем 7-5*d+8d=13 3*d=6 d=2

2*a1=7-5*2 a1=-1,5

S (10) = 10 * (2 * (-1,5) + 2 * (10-1)) / 2=5 * (-3+18) = 75