### Calculate the common ratio, sum of terms of a geometric sequence given the third and seventh term

Question Sample Titled 'Calculate the common ratio, sum of terms of a geometric sequence given the third and seventh term'

Let ${a}_{{n}}$ be the ${n}^{\text{th}}$ term of a geometric sequence. If ${a}_{{3}}={18}$ and ${a}_{{7}}={162}$ , which of the following must be true?

 I. The common ratio of the sequence is less than ${1}$ . II. All of the the terms of the sequence are rational numbers. III. The sum of the first ${99}$ terms of the sequence is greater than ${10}^{{23}}$ .

A
III only
B
II only
C
I and III only
D
II and III only

 I. Let ${r}$ be the common ratio of the geometric sequence.  ${a}_{{3}}\times{r}^{{4}}$ $={a}_{{7}}$ ${18}\times{r}^{{4}}$ $={162}$ ${r}^{{4}}$ $={9}$ ${r}^{{2}}$ $={3}$ or ${r}^{{2}}=-{3}$(rejected) ${r}$ $=\pm\sqrt{{{3}}}$ i.e. ${r}$ $=\sqrt{{{3}}}\approx{1.732}$ or ${r}=-\sqrt{{{3}}}\approx-{1.732}$ I is false. II.  When ${r}=\sqrt{{{3}}}$ , ${a}_{{4}}={a}_{{3}}\times\sqrt{{{3}}}={18}\sqrt{{{3}}}$ Irrational number When ${r}=-\sqrt{{{3}}}$ , ${a}_{{4}}={a}_{{3}}\times-\sqrt{{{3}}}=-{18}\sqrt{{{3}}}$ Irrational number Similarly, all the even terms in the sequence are irrational, i.e. ${a}_{{2}},{a}_{{4}},{a}_{{6}},\ldots$ are irrational, while all the odd terms are rational. I is false. III.  First calculate the first term ${\left({a}_{{1}}\right)}$ of the sequence, ${a}_{{1}}\times{r}^{{2}}$ $={a}_{{3}}$ ${a}_{{1}}\times{3}$ $={18}$ ${a}_{{1}}$ $={6}$ Consider the case of common ratio being negative, i.e. ${r}$ $=-\sqrt{{3}}$ , ${S}{\left({99}\right)}$ $=\dfrac{{{6}{\left({1}-{\left(-\sqrt{{3}}\right)}^{{99}}\right)}}}{{{1}-{\left(-\sqrt{{3}}\right)}}}$ ${S}{\left({n}\right)}=\dfrac{{{a}{\left({1}-{r}^{{n}}\right)}}}{{{1}-{r}}}$ $=\dfrac{{{6}{\left({1}+{3}^{{\tfrac{{99}}{{2}}}}\right)}}}{{{1}+\sqrt{{3}}}}$ ${1}-{\left(-\sqrt{{3}}\right)}^{{99}}={1}-{\left(-{3}^{{\tfrac{{1}}{{2}}}}\right)}^{{99}}={1}-{\left(-{\left({3}^{{\tfrac{{99}}{{2}}}}\right)}\right)}={1}+{\left({3}^{{\tfrac{{99}}{{2}}}}\right)}$ $={9.102581737418128}\times{10}^{{23}}$ $\gt{10}^{{23}}$ ∴   The sum must be greater than ${10}^{{23}}$ III is true.  For your reference, when ${r}=\sqrt{{3}}$, the statement is also true. ${S}{\left({99}\right)}$ $=\dfrac{{{6}{\left(\sqrt{{3}}^{{99}}-{1}\right)}}}{{{\left(\sqrt{{3}}\right)}-{1}}}$ $={3.397129752409304}\times{10}^{{24}}$ $\gt{10}^{{23}}$

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