Question

помогите пожалуйста даю 25 баллов

Answer 1

1.

а)

$a_1 = 16;\ \ \ \ \ \ \ \ a_8 = 37\\\\\\a_8 = a_1 + d(8-1) = a_1 + 7d\\\\d = \dfrac{a_8 - a_1}{7} = \dfrac{37-16}{7} = \dfrac{21}{7} = \bf{3}$

б)

$a_1 = 6;\ \ \ \ \ \ \ \ \ \ a_{18} = -11\\\\\\a_{18} = a_1 + d(18-1) = a_1 + 17d\\\\d = \dfrac{a_{18} - a_1}{17} = \dfrac{-11 - 6}{17} = \dfrac{-17}{17} = \bf{-1}$

2.

$a_2 = 2,\ \ \ \ a_4 = 6\\\\\\\\a_3 = \dfrac{a_4 + a_2}{2} = \dfrac{6+2}{2} = \dfrac{8}{2} = 4\\\\\\d = a_3 - a_2 = 4-2 = 2\\\\a_6 = a_4 + 2d = 6 + 2\cdot 2 = 6 + 4 = \bf{10}$

3.

$d = 6,\ \ \ \ S_{10}= 340\\\\\\\\S_{10} = \dfrac{a_1 + a_{10}}{2} \cdot 10 = 5(a_1 + a_{10})\\\\\\a_1 + a_{10} = \dfrac{S_{10}}{5} = \dfrac{340}{5} = 68\\\\a_{10} = a_1 + d(10-1) = a_1 + 9d\\\\a_1 + a_1 + 9d = 68\\\\2a_1 = 68 - 9d\\\\a_1 = \dfrac{68-9d}{2} = \dfrac{68-9\cdot 6}{2} = \dfrac{68-54}{2} = \dfrac{14}{2} = \bf{7}$

$a_{10} = a_1 + 9d = 7 + 54 = \bf{61}$

4.

$a_1 = 12,\ \ \ \ a_4 = 18\\\\\\\\a_4 = a_1 + d(4-1) = a_1 + 3d\\\\a_1 + 3d = 18\\\\d = \dfrac{18-a_1}{3} = \dfrac{18-12}{3} = \dfrac{6}{3} = 2\\\\\\S_{6} = \dfrac{2a_1 + d(6-1)}{2}\cdot 6 = 3(2a_1 + 5d) = 3(2\cdot 12 + 5\cdot 2) = 3(24 + 10) =\\\\=3\cdot 34 = \bf{102}$

5.

$a_1 = 9,\ \ \ \ \ a_{26} = 44\\\\\\\\a_{26} = a_1 + d(26-1) = a_1 + 25d\\\\a_1 + 25d = 44\\\\d = \dfrac{44-a_1}{25} = \dfrac{44 - 9}{25} = \dfrac{35}{25} = \dfrac{7}{5} = 1,4\\\\a_{15} = a_1 + d(15-1) = a_1 + 14d = 9 + 14\cdot 1,4 = 9 + 19,6 = 28,6\\\\a_{30} = a_1 + d(30-1) = a_1 + 29d = 9 + 29\cdot 1,4 = 9 + 40,6 = 49,6\\\\\\S_{15-30} = \dfrac{a_{15} + a_{30}}{2} \cdot 16 = 8(a_{15} + a_{30}) = 8\cdot (28,6 + 49,6) = 8\cdot 78,2 =\bf{625,6}$

Домножаем на 16, поскольку с 15-го члена по 30-й включительно - всего 16 членов.