The 19 th term of the sequence is the first term of the sequence having value greater than 1000.
Given that the n-th term of the sequence is given by,
[tex]a_n=3n^2[/tex]
Let the p-th term be the first term which have value greater than 1000 in the sequence.
The p-th term of the sequence is, [tex]a_p=3p^2[/tex]
According to the condition,
[tex]3p^2 > 1000[/tex]
[tex]\Rightarrow p^2 > \frac{1000}{3}[/tex]
[tex]\Rightarrow p > +\sqrt{\frac{1000}{3}}[/tex], as number of term cannot be negative so we ignore the negative value of square root here
[tex]\Rightarrow p > 18.25\\\Rightarrow p\geq19[/tex]
Hence the first term which has value greater than 1000 is 19 th term of the sequence.
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