geometric sequence

**polymerase** · June 20th 2007, 08:43 AM

Find a, r, and Tn for each geometric sequence
a)t4=24 and t6=96 b)t2=-6 and t5=-162
-thank you for whoever answers my question

.......

**topsquark** · June 20th 2007, 09:03 AM

Originally Posted by polymerase

Find a, r, and Tn for each geometric sequence
a)t4=24 and t6=96

a) $t_4 =24$ and $t_6 = 96$

In general
$t_n = t_1 \cdot r^{n - 1}$

We have
$t_4 = t_1 \cdot r^3 = 24$
$t_6 = t_1 \cdot r^5 = 96$

Thus
$\frac{t_6}{t_4} = \frac{96}{24} = r^2$

or

$r^2 = 4$

Thus $r = \pm 2$

Take r = 2:
$t_4 = 24 = t_1 \cdot (2)^3 = 8t_1$

Thus $t_1 = 3$ and $t_n = 3(2)^{n - 1}$

Take r = -2:
$t_4 = 24 = t_1 \cdot (-2)^3 = -8t_1$

Thus $t_1 = -3$ and $t_n = -3(-2)^{n - 1}$

Both series will produce the correct terms.

-Dan

**topsquark** · June 20th 2007, 09:06 AM

Originally Posted by polymerase

Find a, r, and Tn for each geometric sequence
b)t2=-6 and t5=-162

Follow the same steps as before.
$t_n = t_1 \cdot r^{n - 1}$

with
$t_2 = -6 = t_1 \cdot r$
$t_5 = -162 = t_1 \cdot r^4$

Dividing I get that
$r = 3$

and from there I get that
$t_1 = -2$

Thus
$t_n = -2(3)^{n - 1}$

-Dan

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