$a, 1, b$ are three consecutive terms of an arithmetic series, find $a, b$ and $S$ of the infinite geometric series

Question

$a, 1, b$ are three consecutive terms of an arithmetic series, and $b, a$ and $\frac{8}{3}$ are the first three terms of an infinite geometric series that has a sum of $S$. find $a, b$ and $S$.

How do I solve this?

Answer 1

by $a, 1, b$ are three consecutive terms of an arithmetic series we can know that:$$a+b=2$$ by $b, a$ and $\frac{8}{3}$ are the first three terms of an infinite geometric series we can know that:$$\frac{8}{3}b=a^2$$ so we got such an equation set:$$\begin{cases}a+b=2\\\frac{8}{3}b=a^2\\\end{cases}$$ the solution of above equation set is:$\begin{cases}a_1=\frac43\\b_1=\frac23\end{cases}$ or $\begin{cases}a_2=-4\\b_2=6\end{cases}$

The first case

the geometric ratio of this geometric sequence $q_1=2$ and $S$ is infinite in this case

The second case

the geometric ratio of this geometric sequence $q_2=-\frac23$ and $S=\lim_{n\to +\infty}\frac{b_2(1-q_2^n)}{1-q_2}=\lim_{n\to +\infty}\frac{6(1-({-\frac23})^n)}{1-(-\frac23)}=\frac{18}{5}$

Answer 2

Since A, 1 and b are in arithmetic progression, so :

$ 1-a = b-1 $, or $a+b = 2 $

Now, since b,a and $ \frac{8}{3} $ are in geometric progression, so: $$a^2 = \frac{8b}{3}$$ $$(2-b)^2 = \frac{8b}{3} $$ $$ 3b^2 - 20b + 12 = 0 $$ $$ b = \frac {20 ^+_- \sqrt{(20^2 - 4.3.12)}} {2.3} $$ $$ b = \frac {20 ^+_- \sqrt{256}} { 6} $$ $$ b = \frac {20 ^+_- 16} {6} $$

$ b = 6 $ or $\frac{2}{3} $

$a = 2 - b$ $ = -4 $ or $ \frac{4}{3} $

There is one important thing to note here. Out of the two sets of values, one set of a and b does not satisfy the arithmetic progression. So you can discard that set of values, hence you have only one unique solution each for a and b.

Now, use the formula for sum of infinite series: $S = \frac{a}{(1-r)}$ , where r = common ratio $= \frac{a}{b} $. I hope you can find that.