Question

an=17+2n
s12-? s5-13-?

Answer 1

A1=19

A5=17+2*5=27

A12=17+2*12=41

s12=(a1+a12)12/2=360

S5-13=(a1+a5)5/2-13=102

Answer 2

$a_n=17+2n\a_1=17+2cdot1=19\ a_4=17+2cdot4=25\a_{12}=17+2cdot12=41\a_{13}=17+2cdot13=43\ \ S_{12}=frac{(a_1+a_{12})cdot12}{2} =(19+41)cdot6= 360\\ S_{4}=frac{(a_1+a_{4})cdot4}{2} =(19+25)cdot2=44cdot2=88\ \S_{13}=S_{12}+a_{13}=360+43=403\ \ S_{5-13}=S_{13}-S_4=403-88=315\ \ OTBET:S_{12}=360;S_{5-13}=315$

Answer 3

Хорошо, сейчас подправлю ;)