# The sum to infinity if a GP is 16 and the sum of the first 4 terms is 15. Find the first four terms?

Jun 1, 2018

The first four terms may either be

$8 , 4 , 2 , 1$

OR

$24 , - 12 , 6 , - 3$

#### Explanation:

We know that the sum of an infinite geometric series is

${s}_{n} = \frac{a}{1 - r}$

The question tells us that ${s}_{n} = 16$.

$16 = \frac{a}{1 - r} \to 16 \left(1 - r\right) = a$

Next we recall that the sum of the first n terms of a geometric progression is

${s}_{N} = \frac{a \left(1 - {r}^{n}\right)}{1 - r}$

$15 = \frac{a \left(1 - {r}^{4}\right)}{1 - r}$

We can simplify the equation a little before combining it with the other one.

$15 = \frac{a \left(1 - {r}^{2}\right) \left(1 + {r}^{2}\right)}{1 - r}$

$15 = \frac{a \left(1 + r\right) \left(1 - r\right) \left(1 + {r}^{2}\right)}{1 - r}$

$\frac{15}{\left(1 + r\right) \left({r}^{2} + 1\right)} = a$

We can now see that

$16 \left(1 - r\right) = \frac{15}{\left(1 + r\right) \left({r}^{2} + 1\right)}$

$16 \left(1 - r\right) \left({r}^{3} + {r}^{2} + r + 1\right) = 15$

$16 \left({r}^{3} + {r}^{2} + r + 1 - {r}^{4} - {r}^{3} - {r}^{2} - r\right) = 15$

$16 - 16 {r}^{4} = 15$

$1 = 16 {r}^{4}$

$\frac{1}{16} = {r}^{4}$

$r = \pm \frac{1}{2}$

We have two possible situations here.

When $r = \frac{1}{2}$

$16 = \frac{a}{1 - \frac{1}{2}} \to a = 16 \left(\frac{1}{2}\right) = 8$

The first four terms here are

$8 , 4 , 2 , 1$

When $r = - \frac{1}{2}$

$16 = \frac{a}{\frac{3}{2}} \to a = 16 \left(\frac{3}{2}\right) = 24$

The first four terms here are

$24 , - 12 , 6 , - 3$

Hopefully this helps!