The $n^{t h}$ term of a sequence is $3 n - 2$ . Is the sequence an A.P. ? If so, find its $10^{t h}$ term.

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Solution

We have, $a_{n} = 3 n - 2$

Clearly an is a linear expression in $n$ . So, the given sequence is an A.P. with common difference $3$ .
(or)
$a_{n} = 3 n - 2$
$\Rightarrow a_{1} = 3 (1) - 2 = 1$
$a_{2} = 3 (2) - 2 = 4$
$a_{3} = 3 (3) - 2 = 7$
$a_{3} - a_{2} = 7 - 4 = 3 a_{2} - a_{1} = 4 - 1 = 3$
i.e., the difference of consecutive terms is constant.
So, $a_{n} = 3 n - 2$ will form an A.P.

Putting $n = 10$ , we get

$a_{10} = 3 \times 10 - 2 = 28$
Therefore, $10^{t h}$ term is $28$ .

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