Дана геометрическая прогрессия:

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/ b1 + b4 = 35/3

/ b2+b3=10

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Найти S5

[b1+b4=35/3⇒b1 (1+q³) = 35/3⇒b1=35/[3 (1+q³) ]

{b2+b3=10⇒b1q (1+q) = 10⇒b1=10/[q (1+q) ]

35/[3 (1+q) (1-q+q²) ]=10/[q (1+q)

7/[3 (1-q+q²) ]=2/q

6 (1-q+q²) = 7q

6-6q+6q²-7q=0

6q²-13q+6=0

D=169-144=25

q1 = (13-5) / 12=2/3⇒b1=10: (2/3+4/9) = 10:10/9=10*9/10=9

S5=b1 * (1-q^5) / (1-q)

S5=9 * (1-32/243) : (1-2/3) = 9*211/243*3=211/9=23 4/9

q2 = (13+5) / 12=3/2⇒b1=10: (3/2+9/4) = 10:15/4=10*4/15=8/3

S5=8/3 * (243/32-1) : (3/2-1) = 8/3*211/32*2=211/6=35 1/6