geometric series/sequence given question

**johnsy123** · July 12th 2010, 01:35 AM

consider the following geometric sequences and find the terms indicated:

*The 1st term is 440 and the 12th is 880. find S6

**Prove It** · July 12th 2010, 01:41 AM

You should know that $t_n = ar^{n-1}$ .

So $a = t_1 = 440$ and $t_{12} = ar^{11} = 880$ .

This means $440r^{11} = 880$

$r^{11} = 2$

$r =2^{\frac{1}{11}}$ .

This means $t_n = 440(2^{\frac{1}{11}})^{n-1}$ .

You should also know that $S_n = \frac{a(r^n-1)}{r-1}$

So $S_6 = \frac{440[(2^{\frac{1}{11}})^6-1]}{2^{\frac{1}{11}} - 1}$

$= \frac{440(2^{\frac{6}{11}} - 1)}{2^{\frac{1}{11}} - 1}$ .

Use a calculator to simplify.

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