В арифметической прогрессии (аn), найти d, a, S7 если а3=16; а

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3a^2+5ab=19+2b^2

a, b > 0 a,b ∈ N

разложим 3a^2+5ab=2b^2

3a^2+5ab -2b^2 = 0

D = (5b)^2 + 4*3*2b^2 = 25b^2 + 24b^2 = 49b^2

a12 = (-5b +- 7b)/6 = -2b   1/3b

3a^2+5ab -2b^2 = (a - 1/3b)(a + 2b) = (3a - b)(a + 2b)

получили  

(3a - b)(a + 2b) = 19

19 простое делится на +- 1 и +- 19

значит и множители могут быть только целыми в левой части

19 = 1*19 = (-1) * (-19)

получаем системы

1. 3a - b = -1

a + 2b = -19   нет a, b > 0

2. 3a - b = -19

a + 2b = -1    нет a, b > 0

3. 3a - b = 1

a + 2b = 19

4. 3a - b = 19

a + 2b = 1     нет a, b > 0

решаем только одну систему

3a - b = 1

a + 2b = 19

--

b = 3a - 1

a + 2(3a - 1) = 19

a + 6a -2 = 19

7a = 21

a = 3

b = 3a - 1 = 3*3 - 1 = 8

ответ (3, 8)