Question

6с.50 6а.22 6б.38 под буквами б пожалуйста)

Answer 1

6А.22б
$S_n= \frac{b_1(q^n-1)}{q-1} \\\ S_7= \frac{b_1(q^7-1)}{q-1} \\\ b_5=b_1q^4 \\\ b_1= \frac{b_5}{q^4} = \cfrac{ \frac{16}{9} }{(\frac{2}{3} )^4} = \cfrac{ \frac{16}{9} }{\frac{16}{81} } =9 \\\ S_7= \frac{9\cdot(( \frac{2}{3} )^7-1)}{\frac{2}{3}-1} = \frac{9\cdot( \frac{128}{2187} -1)}{-\frac{1}{3}} = \frac{9\cdot(- \frac{2059}{2187})}{-\frac{1}{3}} =3^3\cdot \frac{2059}{3^7}= \frac{2059}{81}=25 \frac{34}{81}$

6C.50б
$S_n= \frac{b_1(q^n-1)}{q-1} \\\ S_5= \frac{b_1(q^5-1)}{q-1} \\\ b_5=b_1q^4 \\\ b_1= \frac{b_5}{q^4} = \cfrac{1\frac{1}{9} }{(\frac{1}{3} )^4} = \cfrac{ \frac{10}{9} }{\frac{1}{81} } =90 \\\ S_5= \frac{90\cdot (( \frac{1}{3} )^5-1)}{ \frac{1}{3} -1} = \frac{90\cdot (\frac{1}{243} -1)}{ -\frac{2}{3} } = \frac{90\cdot (-\frac{242}{243} )}{ -\frac{2}{3} } = 90\cdot \frac{121}{81}= \frac{1210}{9}=134 \frac{4}{9}$

6В.38б
$S_n= \frac{b_1(q^n-1)}{q-1} \\\ 15 \sqrt{10}+15 \sqrt{5} = \frac{ \sqrt{5} (( \sqrt{2} )^n-1)}{ \sqrt{2} -1} \\\ 15 \sqrt{2}+15 = \frac{( \sqrt{2} )^n-1}{ \sqrt{2} -1} \\\ 15( \sqrt{2}+1) = \frac{( \sqrt{2} )^n-1}{ \sqrt{2} -1} \\\ 15( \sqrt{2}+1)(\sqrt{2} -1) =( \sqrt{2} )^n-1 \\\ 15(2 -1) =( \sqrt{2} )^n-1 \\\ 15 =( \sqrt{2} )^n-1 \\\ ( \sqrt{2} )^n=16 \\\ ( \sqrt{2} )^n=( \sqrt{2} )^8 \\\ n=8$