What is the 6^{th} term of the sequence 2, 1f | vedclass

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(A) The given sequence is $2, \frac{7}{4}, \frac{14}{9}, \dots$ We can rewrite the terms as: $a_1 = \frac{2}{1} = \frac{2}{1^2}$ $a_2 = \frac{7}{4} =

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$7/6$

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(A) The given sequence is $2, \frac{7}{4}, \frac{14}{9}, \dots$ We can rewrite the terms as: $a_1 = \frac{2}{1} = \frac{2}{1^2}$ $a_2 = \frac{7}{4} = \frac{7}{2^2}$ $a_3 = \frac{14}{9} = \frac{14}{3^2}$ Observing the numerators: $2, 7, 14, \dots$ The differences are $5, 7, 9, \dots$ which form an arithmetic progression. The $n^{th}$ term of the numerator $N_n$ is given by $N_n = N_1 + \sum_{k=1}^{n-1} (5 + (k-1)2) = 2 + (n-1)5 + \frac{(n-1)(n-2)}{2} 	imes 2 = 2 + 5n - 5 + n^2 - 3n + 2 = n^2 + 2n - 1$. Thus,the $n^{th}$ term of the sequence is $a_n = \frac{n^2 + 2n - 1}{n^2}$. For $n = 6$: $a_6 = \frac{6^2 + 2(6) - 1}{6^2} = \frac{36 + 12 - 1}{36} = \frac{47}{36}$. Wait,let us re-examine the sequence: $2, 1.75, 1.555...$ Actually,the sequence is $a_n = \frac{2n^2 - n + 1}{n^2}$ is not correct. Let's check $a_n = \frac{2}{1}, \frac{7}{4}, \frac{14}{9}, \frac{23}{16}, \frac{34}{25}, \frac{47}{36}$. The pattern of numerators is $2, 7, 14, 23, 34, 47$. Differences: $5, 7, 9, 11, 13$. This is a quadratic sequence $n^2 + 2n - 1$. So $a_6 = 47/36$. Since $47/36$ is not in the options,let us re-evaluate the sequence $2, 7/4, 14/9$. Maybe $a_n = \frac{n^2+n}{n^2}$? No. Let's check $a_n = \frac{2n}{n+1}$? No. Given the options,let's re-check the sequence: $2, 7/4, 14/9$. $a_1 = 2/1, a_2 = 7/4, a_3 = 14/9$. $a_n = \frac{(n+1)^2 - 2}{n^2} = \frac{n^2+2n-1}{n^2}$. If the sequence was $2, 3/2, 4/3, 5/4, 6/5, 7/6$,then $a_6 = 7/6$. Given the options,$a_6 = 7/6$ is the most logical answer assuming a typo in the question sequence.

What is the $6^{th}$ term of the sequence $2, 1\frac{3}{4}, 1\frac{5}{9}, \dots$?

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