Question

How do you find the first five terms in the geometric sequence which is such that the sum of the 1st and 3rd terms is 50, and the sum of the 2nd and 4th terms is 150?

Answer 1

a + ar^2 = 50 (1)

ar + ar^3 = 150 (2)

divide (2) by (1)

$\frac{ar + a {r}^{3} }{a + {ar}^{2} } = \frac{150}{50} \\ \frac{ar(1 + {r}^{2} )}{a(1 + {r}^{2}) } = \frac{150}{50} \\ r = 3 \\ substitute \: r \: in \: (1) \\ a + a {r}^{2} = 50 \\ a(1 + {r}^{2} ) = 50 \\ a(1 + 9) = 50 \\ a \times 10 = 50 \\ a = \frac{50}{10} = 5 \\ the \: terms \: are \\ a \: ar \: a {r}^{2} \: a {r}^{3} \: a {r}^{4} \\ 5 \: \: 15 \: \: 45 \: \: 135 \: \: 405$

5, 15, 45, 135, 405