Suku kedua suatu deret geometri 3/2 dan jumlah deret sampai tak hingga ialah 6 tentukan suku pertama dan rasionya

[tex]u2 = \frac{3}{2} \\ ar = \frac{3}{2} \\ \\ {s}^{ \infty } = \frac{a}{1 - r} \\ 6 = \frac{a}{1 - r} \\ 6 - 6r = a \\ \\ substitusi \\ (6 - 6r)r = \frac{3}{2} \\ 2(6r - 6 {r}^{2} ) = 3 \\ 12r - 12 {r}^{2} - 3 = 0 \\ 4 {r}^{2} - 4r + 1 = 0 \\ (r = \frac{1}{2} ) \\ \\ ar = \frac{3}{2} \\ a = \frac{3}{2r} \\ a = \frac{3}{2 \times \frac{1}{2} } \\ a = 3[/tex]

Stak hingga = a/(1 - r)
6 = a/(1 - r)
6 × (1 - r) = a
6 - 6r = a

U2 = a × r^(2-1)
3/2 = (6 - 6r) × r
3/2 = 6r - 6r^2
6r^2 - 6r + 3/2 = 0
pq = 9
p + q = -6
p = -3
q = -3

6r^2 - 6r - 3/2 = 0
6r^2 - 3r - 3r - 3/2 = 0
3r(2r - 1) + 3/2(2r - 1) = 0
(3r + 3/2) (2r - 1) = 0
3r + 3/2 = 0
3r = -3/2
r1 = -1/2
atau
2r - 1 = 0
2r = 1
r = 1/2
r1, r2 = -1/2, 1/2

a = 6 -6r
= 6 - (6 × 1/2)
= 6 - 3
= 3

Yang paling mungkin r = 1/2 dan a = 3.

Semoga membantu ^^