Let the first term and the common ratio r of a geometric progression be positive integers.

Question

Let the first term $\alpha $ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Accepted Answer

Given that the first term $a$ and common ratio $r$ of a geometric progression be positive integer. So, their 1st three terms are $a,ar,a{r}^{2}$

According to the question, ${a}^{2}+{a}^{2}{r}^{2}+{a}^{2}{r}^{4}=33033$

$\begin{gathered} \Rightarrow {a}^{2}(1+{r}^{2}+{r}^{4}) =3	imes 7	imes 11	imes 11	imes 13 \ =3	imes 7	imes 13	imes {11}^{2} \end{gathered}$

$\begin{gathered} 	herefore \, {a}^{2}={11}^{2} \ \Rightarrow \, a=11 \ 1+{r}^{2}+{r}^{4}=273 \ \Rightarrow {r}^{2}+{r}^{4}=272 \ \Rightarrow {r}^{4}+{r}^{2}-272=0 \ \Rightarrow ({r}^{2}+17)({r}^{2}-16)=0 \ \Rightarrow {r}^{2}=-17(notpossible), \ \Rightarrow {r}^{2}-16=0 \ \Rightarrow r=\pm 4 \ \Rightarrow r=4(\because r>0) \end{gathered}$

So, sum of these first three terms is $a+ar+a{r}^{2}$ $=11+44+176=231$

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