Question

va rog frumosss ajutati ma si pe mibe E URGENT, imi trebyie repede ​

Answer 1

$\displaystyle 3.~~a_1=-10,~r=4\\\\ a_5,~a_8,~S_{10}=?\\\\a_n=a_1+(n-1)\cdot r\\\\ a_5=a_1+4r=-10+4 \cdot 4=-10+16=6\\\\ a_8=a_1+7r=-10+7 \cdot 4=-10+28=18\\\\ S_n=\frac{n[2a_1+(n-1)\cdot r]}{2} \\\\S_{10}=\frac{10[2 \cdot (-10)+(10-1)\cdot 4}{2} =\frac{10(-20+9 \cdot 4)}{2} =\frac{10(-20+36)}{2} =\\\\=\frac{10 \cdot 16}{2} =\frac{160}{2} =80$

$\displaystyle 4.~b_1=-128,~q=\frac{1}{2} \\\\ b_4,~b_6,~b_8=?\\\\ b_n=b_1 \cdot q^{n-1}\\\\ b_4=-128\cdot \Bigg(\frac{1}{2} \Bigg)^{4-1}=-128 \cdot \Bigg(\frac{1}{2} \Bigg)^3=-128 \cdot \frac{1^3}{2^3} =-128 \cdot \frac{1}{8} = -16\\\\ b_6=-128 \cdot \Bigg(\frac{1}{2} \Bigg)^{6-1}=-128\cdot \Bigg(\frac{1}{2} \Bigg)^5=-128 \cdot \frac{1^5}{2^5} =-128 \cdot \frac{1}{32}=-4\\\\b_8=-128 \cdot\Bigg(\frac{1}{2}\Bigg)^{8-1}=-128\cdot\Bigg(\frac{1}{2}\Bigg)^7=-128\cdot\frac{1^7}{2^7}= -128\cdot\frac{1}{128}=-1$