A geometric series has a first term $\sqrt{2}$ and a second term $\sqrt{6}$ . Find the 12th term and the sum of the first 12 terms.

I can get to the answers as irrational numbers using a calculator but how can I can obtain the two answers in radical form $243 * \sqrt{6}$ and $364 \left(\sqrt{6}+\sqrt{2}\right)$ ?

The closest I get with the 12th term is $\sqrt{2} \left(\sqrt{6} \over \sqrt{2}\right)^{(12-1)}$ or $\sqrt{2} * 3^\left({11\over 2}\right)$

And for the sum ${\sqrt{2}-\sqrt{2}*(\sqrt{3})^{12} \over 1 - \sqrt{3}}$

  • 1
    $\begingroup$ try to use $$ to make it readable $\endgroup$ – Nikita Evseev Aug 15 '13 at 14:53

The general term of a geometric series is $a r^{n-1}$, where $r$ is the ratio, and $a$ is the first term. In your case, $a=\sqrt{2}$ and $r=\sqrt{3}$. The 12th term is then

$$\sqrt{2} (\sqrt{3})^{11} = 243 \sqrt{6}$$

The sum of the first 12 terms is

$$a \sum_{k=0}^{11} r^k = a \frac{r^{12}-1}{r-1} = \sqrt{2} \frac{728}{\sqrt{3}-1} = 364 (\sqrt{2}+\sqrt{6})$$

Note that I used $3^5 = 243$ and $3^6=729$ in the above.


Note that

$$ \sqrt{2} \frac{728}{\sqrt{3}-1} = \sqrt{2} \frac{728}{\sqrt{3}-1} \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{2} (\sqrt{3}+1) 728}{(\sqrt{3})^2-1^2} = \frac{(\sqrt{6}+\sqrt{2}) 728}{3-1}$$

  • $\begingroup$ Please can you expand how you got from $\sqrt{2} \frac{728}{\sqrt{3}-1}$ to $364 (\sqrt{2}+\sqrt{6})$ ? $\endgroup$ – Gez Bishop Aug 15 '13 at 15:08
  • 1
    $\begingroup$ @GezBishop: see edit. $\endgroup$ – Ron Gordon Aug 15 '13 at 15:12

So, the common ratio $=\frac{\sqrt6}{\sqrt2}=\sqrt3$

So, the $n$ th term $=\sqrt2(\sqrt3)^{n-1}\implies 12$th term $=\sqrt2(\sqrt3)^{12-1}=\sqrt2(\sqrt3)^{11}$

Now, $\displaystyle(\sqrt3)^{11}=\sqrt3 \cdot 3^5=243\sqrt3$

The sum of $n$ term is $\displaystyle \sqrt2\cdot\frac{(\sqrt3)^n-1}{\sqrt3-1}$

$\implies 12$th term $=\displaystyle \sqrt2\cdot\frac{(\sqrt3)^{12}-1}{\sqrt3-1}=\sqrt2\cdot\frac{(3^6-1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}$ (rationalizing the denominator )



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.