# If the first three terms of a Geometric Series are (k-8),(k+4) and (3k+2), what is (a) the value of k; (b) what is the sixth term of the series and (c) what is the sum of first 10 terms?

Jan 14, 2018

$k = 16$ or $k = - 1$ - hence two solutions. See details below.

#### Explanation:

As $\left(k - 8\right)$, $\left(k + 4\right)$ and $\left(3 k + 2\right)$ are in geometric series

${\left(k + 4\right)}^{2} = \left(k - 8\right) \left(3 k + 2\right)$

or ${k}^{2} + 8 k + 16 = 3 {k}^{2} - 22 k - 16$

or $2 {k}^{2} - 30 k - 32 = 0$

or ${k}^{2} - 15 k - 16 = 0$

or $\left(k - 16\right) \left(k + 1\right) = 0$

i.e. $k = 16$ or $k = - 1$

Hence, theuir could be two geometric series

(1) If $k = 16$, first three terms are $8 , 20$ and $50$ i.e. first term is $8$ and common ratio is $\frac{20}{8} = \frac{5}{2}$ and sixth term is $8 \cdot {\left(\frac{5}{2}\right)}^{5} = \frac{3125}{4}$ and sum of first $10$ terms is $8 \frac{{\left(\frac{5}{2}\right)}^{10} - 1}{\frac{5}{2} - 1} = 8 \left(\frac{9765625}{1024} - 1\right) \times \frac{2}{3} = 8 \times \frac{9764601}{1024} \times \frac{2}{3} = \frac{3254867}{64} = 50857 \frac{19}{64}$

(2) If $k = - 1$, first three terms are $- 9 , 3$ and $- 1$ i.e. first term is $- 9$ and common ratio is $\frac{3}{- 9} = - \frac{1}{3}$ and sixth term is $\left(- 9\right) \cdot {\left(- \frac{1}{3}\right)}^{5} = \frac{1}{27}$ and sum of first $10$ terms is $\left(- 9\right) \frac{{\left(- \frac{1}{3}\right)}^{10} - 1}{- \frac{1}{3} - 1} = 9 \left(\frac{1}{59049} - 1\right) \times \frac{3}{4} = - \frac{27}{4} \times \frac{59048}{59049} = - \frac{14762}{2187} = - 6 \frac{1640}{2187}$