# On a geometry sequence, total the 4th and the 6th terms equals 30. While the 3rd term equals 3. The 8th term equals?

Jan 21, 2018

${8}^{t h}$ term is $96$

#### Explanation:

Let the first term of geometric sequence be ${a}_{1}$ (${a}_{n}$ indicating ${n}^{t h}$ term) and common ratio be $r$. Then ${a}_{3} = {a}_{1} {r}^{2}$, ${a}_{4} = {a}_{1} {r}^{3}$, ${a}_{6} = {a}_{1} {r}^{5}$ and ${a}_{8} = {a}_{1} {r}^{7}$.

As ${a}_{3} = 3$, we have ${a}_{1} {r}^{2} = 3$ or ${a}_{1} = \frac{3}{r} ^ 2$

andas total of ${4}^{t h}$ and the ${6}^{t h}$ terms equals $30$, we have

${a}_{1} {r}^{3} + {a}_{1} {r}^{5} = 30$

i.e. ${a}_{1} {r}^{2} \left(r + {r}^{3}\right) = 30$

or $3 \left(r + {r}^{3}\right) = 30$

or ${r}^{3} + r = 10$ or ${r}^{3} + r - 10 = 0$

i.e. $\left(r - 2\right) \left({r}^{2} - 2 r + 5\right) = 0$

or $\left(r - 2\right) \left({\left(r - 1\right)}^{2} + 4\right) = 0$

as ${\left(r - 1\right)}^{2} + 4$ is always positive, we have $r - 2 = 0$ or $r = 2$

Hence ${a}_{1} = \frac{3}{4}$ and ${a}_{8} = {a}_{1} {r}^{7} = \frac{3}{4} \times {2}^{7} = 96$

i.e. ${8}^{t h}$ term is $96$.