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

n => a^2+a^2 r^2+a^2 r^4=33033 n => a^2( r^4+ r^2+1)=3 x 7 x 11^2 x 13 n => a=11 n => r^4+ r^2+1=273 n => r^4+ r^2-272=0 n => ( r^2+17)( r^2-16)=0 n => r^2=16 => r= ± 4 n t_1+ t_2+ t_3=a+a r+a r^2 n =11+44+176=231nn n

Let the first term a and the common ratio r of a geomet... | Sequence & Series (Mathematics)

\begin{aligned}\n& \Rightarrow a^2+a^2 \mathrm{r}^2+a^2 \mathrm{r}^4=33033 \\\n& \Rightarrow a^2\left(\mathrm{r}^4+\mathrm{r}^2+1\right)=3 \times 7 \times 11^2 \times 13 \\\n& \Rightarrow a=11 \\\n& \Rightarrow \mathrm{r}^4+\mathrm{r}^2+1=273 \\\n& \Rightarrow \mathrm{r}^4+\mathrm{r}^2-272=0 \\\n& \Rightarrow\left(\mathrm{r}^2+17\right)\left(\mathrm{r}^2-16\right)=0 \\\n& \Rightarrow \mathrm{r}^2=16 \Rightarrow \mathrm{r}= \pm 4 \\\n& \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=a+a \mathrm{r}+a \mathrm{r}^2 \\\n& =11+44+176=231\n\n\end{aligned}\n

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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