Which of the following can be the $n^{t h}$ term of an AP?

2 n^{2} - 3

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6 n^{2} + 1

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1 + n + n^{2}

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5 n + 5

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Solution

The correct option is D $5 n + 5$
Consider: $a_{n} = 2 n^{2} - 3$
$a_{1} = 2 - 3 = - 1$
$a_{2} = 2 \times 22 - 3 = 8 - 3 = 5$
$a_{3} = 2 \times 32 - 3 = 2 \times 9 - 3 = 18 - 3 = 15$
Now, $d_{1} = a_{2} - a_{1} = 5 - (- 1) = 6$
$d_{2} = a_{3} - a_{2} = 15 - 5 = 10$
Here, $d_{1} \neq d_{2}$
Thus, this is not an AP.
Consider: $a_{n} = 6 n^{2} + 1$
So, $a_{1} = 6 \times 1^{2} + 1 = 6 + 1 = 7$
$a_{2} = 6 \times 2^{2} + 1 = 6 \times 4 + 1 = 25$
$a_{3} = 6 \times 3^{2} + 1 = 6 \times 9 + 1 = 55$
Now, $d_{1} = a_{2} - a_{1} = 25 - 7 = 18$
$d_{2} = a_{3} - a_{2} = 55 - 25 = 30$
Here, $d_{1} \neq d_{2}$
Thus, this is not an AP.
Consider: $a_{n} = 1 + n + n^{2}$
So, $a_{1} = 1 + 1 + 1^{2} = 3$
$a_{2} = 1 + 2 + 2^{2} = 1 + 2 + 4 = 7$
$a_{3} = 1 + 3 + 3^{2} = 1 + 3 + 9 = 13$
Now, $d_{1} = a_{2} - a_{1} = 7 - 3 = 4$
$d_{2} = a_{3} - a_{2} = 13 - 7 = 6$
Here, $d 1 \neq d 2$
Thus, this is not an AP.
Consider: $a_{n} = 5 n + 5$
$a_{1} = 5 \times 1 + 5 = 5 + 5 = 10$
$a_{2} = 5 \times 2 + 5 = 10 + 5 = 15$
$a_{3} = 5 \times 3 + 5 = 15 + 5 = 20$
Now, $d_{1} = a_{2} - a_{1} = 15 - 10 = 5$
$d_{2} = a_{3} - a_{2} = 20 - 15 = 15$
Here, $d_{1} = d_{2}$
Thus, this is an AP.
Hence, the correct answer is option (4).

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Similar questions

Which of the following could be the $n^{t h}$ term of any AP?

a_{n}

n^{t h}

n^{t h}

2 n + 3

n^{t h}

n^{2} + 1

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3 n^{2} + 5

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