Question

Hello help me with question b

Answer 1

1. Common ratio = 3
2. First term = 6

Explanation:

`\bf\longrightarrow \large\frac{ \frac{ \frac{a( {r}^(4) - 1)}{ \cancel{r - 1}} }{a( {r}^(2) - 1) } }{ \cancel{r - 1}} = 10 \\`

`\bf\longrightarrow {r}^(4) - 1 = {10r}^(2) - 10`

`\bf\longrightarrow {r}^(4) - {10r}^(2) + 9 = 0`

Put a = r²

`\bf\longrightarrow \: {a}^(2) - 10 {a}^(2) + 9 = 0`

`\bf\longrightarrow \: a = (10± √(100 - 36) )/(2) = (10±8)/(2)`

a = 9 or a = 1

r = 3 or r = 1 but r ≠ 1

`\therefore` r = 3

`\bf\longrightarrow \: a_3 = 54 \\ \bf\longrightarrow \: a {r}^(2) = 54 \\ \bf\longrightarrow \: a = (54)/(9) = 6`