I'm self-studying maths, so have turned to the good people of MSE as I have no teacher to ask.
I completed the following problem and didn't find it difficult. However, the answer provided in the text book is different, and I can't see what I've done wrong. Any guidance would be greatly appreciated.
The Question
$x$, $3$, and $x + 8$ are the fourth, fifth and sixth terms of a geometric series. a) Find two possible values of $x$ and corresponding values of common ratio. b) Given sum to infinity exists, find first term c) Find sum to infinity
My Solution
a) I used the fact of the relationship between consecutive terms in a geometric series.
i.e. $\frac{U2} {U1}$ = $\frac{U3} {U2}$ ... $\frac{U5} {U4}$ = $\frac{U6} {U5}$ So: $\frac{3} {x} = \frac{x + 8} {3}$, so $x = 1$ or $x = -9$
Substituting $x$ into $\frac{U5} {U4}$ in the above: $r = 3$ or $r = \frac {-1} {3}$
Parts b) and c) follow from results in a) so no need to consider them.
The Text Book
$ar^3 = x$
$ar^4 = 3$
$ar^5 = x + 8$
So: $\frac {ar^5} {ar^4} = \frac {ar^4} {ar^3}$
They then express this as $\frac {x + 8} {3} = \frac{3} {x}$, and find the same values for $x$.
However, book says $r = $ $\frac {ar^4} {ar^3}$ $=$ $\frac{x} {3}$ thus giving $r = \frac{1} {3}, r = -3.$ for the found values of $x$
I can't fathom why the method I used is wrong. Many thanks in advance.