I'm having trouble blending two different types of progressions:
The fourth, eighth and fourteenth terms of an A.P., common difference 0.5, are in geometric progression. Find the first term of the A.P. and the common ratio of the G.P.
In a A.P, $u_n = a + (n-1)d$, so:
$$
u_4 = a + \frac{3}{2}, u_8 = a + \frac{7}{2}, u_{14} = a + \frac{13}{2}
$$
but in a geometric series, the difference between $u_n$ and $u_{n+1}$ is $r$:
$$
u_{14} – u_8 = r = \frac{13}{2} – \frac{7}{2} = 3
$$
$$
u_8 – u_4 = r = \frac{7}{2} – \frac{3}{2} = 2
$$
which doesn't make sense, because it looks like $r$ is varying. I also know that the fourth term in the A.P. is offset from the first term by $1.5$ – but I have no means to find $a$. Can somebody point me in the right direction?
Best Answer
In a GP, the ratio is $r$. So $$\frac{u_{14}}{u_8} = r = \frac{u_8}{u_4}$$