Question

арифметическая прогрессия

Answer 1

1.

а) $a_{8}= a_{1}+7d; d=\frac{a_{8}-a_{1}}{7} = \frac{37-16}{7} =\frac{21}{7} =3$

б) $a_{18}= a_{1}+17d; d=\frac{a_{18}-a_{1}}{17} = \frac{-11-6}{17} =\frac{-17}{17} =-1$

2.

$a_2=a_1+d,a_4=a_1+3d\\d=\frac{a_4-a_2}{2} =\frac{6-2}{2} =2\\a_1=a_2-d=2-2=0\\a_6=a_1+5d=0+5*2=10$

3.

$S_{10}=\frac{2a_1+d*(n-1)}{2}*n\\340=\frac{2a_1+6*(10-1)}{2}*10\\2a_1+6*9=34*2\\2a_1=68-54\\2a_1=14\\a_1=7\\a_{10}=a_1+9d=7+9*6=7+54=61$

4.

$a_4=a_1+3d; d=\frac{a_4-a_1}{3}=\frac{18-12}{3} =2\\S_{6}=\frac{2*12+2*(6-1)}{2}*6=(24+10)*3=34*3=102$

5.

$a_{26}=a_1+25d; d=\frac{a_{26}-a_1}{25}=\frac{44-9}{25}=\frac{35}{25} =1.4\\S_{15}=\frac{2*9+1.4*(15-1)}{2}*15=(18+19.6)*7.5=37.6*7.5=282\\S_{30}=\frac{2*9+1.4*(30-1)}{2}*30=(18+40.6)*15=58.6*15=879\\S_{15-30}=S_{30}-S_{15}=879-282=597$