# Question #931d4

Apr 30, 2017

We have to use that the geometric series is of the form:

$a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots$

and that we know that the whole sum is $13.5$ and the sum of the first three terms $a + a r + a {r}^{2} = 13$

#### Explanation:

So we have:

$a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots = 13.5$

and

$a + a r + a {r}^{2} = 13$

Subtracting we have:

$\left(a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots\right) - \left(a + a r + a {r}^{2}\right) = 0.5$, that is:

$a {r}^{3} + a {r}^{4} + a {r}^{5} + a {r}^{6} + a {r}^{7} + \ldots = 0.5$

Hence, taking $r$ as a common factor:

${r}^{3} \cdot \left(a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots\right) = 0.5$

But we know that that the expression between brackets:

$a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots = 13.5$

so we have:

${r}^{3} \cdot 13.5 = 0.5$, that is:

${r}^{3} = \frac{0.5}{13.5}$, that is:

${r}^{3} = \frac{\frac{1}{2}}{\frac{27}{2}} = \frac{1}{27}$, and then

$r = \frac{1}{3}$

But we know that $a + a r + a {r}^{2} = 13$, so:

$a + a \frac{1}{3} + a \frac{1}{9} = 13$, and that means:

$\frac{9 a + 3 a + a}{9} = 13$, so:

$\frac{13 a}{9} = 13$, so $\frac{a}{9} = 1$, and from that we get:

$a = 9$, which is the first term