A3+a7=5, a4=1. Найти сумму первых десяти членов прогрессии Найти сумму целых решений неравенства 0

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A3 = A1 + 2
A7 = A1 + 6d

Given that A3 + A7 = 5
A1 + 2d + A1 + 6d =
2A1 + 8d =
A1 + 4d = 2.5

Given that a4 = 1
A1 + 3d = 1

Subtracting the two equations above
A1 + 4d = 2.
A1 + 3d = 1

d = 1.
A1 = -2

The arithmetic progression would be: -2, -0.5, 1, 2.5, 4, 5.5, 7, 8.5, 10, 11.5

Sum of the first 10 terms = 46

For the inequality 0 < A1 + 9d
-2 + 9*1.5 = 11.5

There are 12 integers between 0 and 11.5, so the sum of all the integers would be 78.