Jika suku ke 3 dan suku ke 5 dari barisan geometri masing-masing 9 dan 1. tentukan U7 dan S7

U3=9, U5=1
a=U1
[tex]u3 = a {r}^{n - 1} [/tex]
=
[tex]a {r}^{3 - 1} [/tex]
=
[tex]a {r}^{2} = 9[/tex]
[tex]u5 = ar {}^{5 - 1} = ar {}^{4} = 1[/tex]
[tex] \frac{u5}{u3} = \frac{1}{9} [/tex]
[tex] \frac{ar^{4} }{ar {}^{2} } = \frac{1}{9} [/tex]

r^2 = 1/9
r = 1/3
U3= ar^2
9 = a.1/3^2
9= a. 1/9
a= 9×9 = 81

U7 = ar^(n-1)
=ar^(7-1)
= ar^6
= 81 . 1/3^6
=81. 1/729
=1/9
S7 = a(1 - r^n) ÷ (1-r)
= 81 ( 1 - 1/3^7) ÷ (1-1/3)
= 81 (1-1/2187) ÷ (2/3)
= 81 × 2186/2187 × 3/2
= 1093/9

[tex]u3 = 9 \\ a {r}^{2} = 9 \\ \\ u5 = 1 \\ {ar}^{4} = 1 \\ {ar}^{2} {r}^{2} = 1 \\ 9 {r}^{2} = 1 \\ {r}^{2} = \frac{1}{9} \\ {r}^{2} = { \frac{1}{3} }^{2} \\ r = \frac{1}{3} \\ \\ {ar}^{2} = 9 \\ a { \frac{1}{3} }^{2} = 9 \\ a = 81[/tex]

[tex]u7 = {ar}^{6} \\ u7 = {81 \times \frac{1}{3} }^{6} \\ u7 = 81 \times \frac{1}{729} \\ u7 = \frac{1}{9} [/tex]

[tex]s7 = \frac{a(1 - {r}^{7}) }{1 - r} \\ s7 = \frac{81(1 - { \frac{1}{3} }^{7}) }{1 - \frac{1}{3} } \\ s7 = \frac{81(1 - \frac{1}{2187} ) }{ \frac{2}{3} } \\ s7 = \frac{81( \frac{2186}{2187} ) }{ \frac{2}{3} } \\ s7 = \frac{2186}{27} \times \frac{3}{2} \\ s7 = \frac{1093}{9} [/tex]