Question

Задание во вложении,очень срочно надо!!!Зделать только осталось 4,5,7 и 8

Answer 1

$4.;a_{12}=-40, ;a_{18}=-22\ a_{12}=a_1+11d\ a_{18}=a_1+17d\ begin{cases} a_1+11d=-40\ a_1+17d=-22 end{cases}Rightarrow begin{cases} a_1=-40-11d\ -40-11d+17d=-22 end{cases}Rightarrow\ begin{cases} a_1=-73\ d=3 end{cases}\ a_n=a_1+(n-1)d=-73+(n-1)3=-73+3n-3=-76+3n\ 5.;a_n<0\ -76+3n<0\ 3n=76\ n<25frac13\ ninmathbb{Z}Rightarrow nleq25\ a_{25}=-76+3cdot25=-1\ a_{26}=-76+3cdot26=2\ S_{25}=frac{a_1+a_{25}}2cdot25=frac{-73-1}2cdot25=925$

$7.;q=-frac13,;b_9=frac4{81}\ b_9=b_1cdot q^8\ frac4{81}=b_1cdotleft(-frac13right)^8\ frac4{81}=b_1cdotfrac1{6561}\ b_1=frac4{81}cdot6561=4cdot81=324$

$8.;b_1=sqrt3,;b_2=-1,;b_3=frac1{sqrt3}\ q=frac{b_2}{b_1}=frac{-1}{sqrt3}=-frac1{sqrt3}\ S_5=frac{b_1(q^5-1)}{q-1}=frac{sqrt3left(-frac1{9sqrt3}-1right)}{-frac1{sqrt3}-1}=frac{frac19+sqrt3}{frac1{sqrt3+1}}=frac{frac19(1+9sqrt3)}{frac1{sqrt3}(1+sqrt3)}=\ =frac{sqrt3(1+9sqrt3)}{9+9sqrt3}$