Question

The nth term of a sequence is in 3n-2 is the sequence an A.P if so find 10th term?​

Answer 1

Answer:

Given that

$a_n = 3n - 2 \\ so \\ a_1 = 3 - 2 = 1\\ a_2 = 3 \times 2 - 2 = 4 \\ a_3 = 3 \times 3 - 2 = 7 \\ a_4 = 3 \times 4 - 2 = 10 \\ a_5 = 3 \times 5 - 2 = 13$

So,

Given sequence will be an AP with

first term = a = 1

and

common difference = d = 3.

So,

10th term of this AP will be

$a_{10} = a + 9d \\ = 1 + 9 \times 3 \\ 1 + 27 \\ = 28$

we can find its 10th term also as

$a_n = 3n - 2 \\ \implies a_{10} = 3 \times 10 - 2 \\ = 30 - 2 \\ = 28$

So 10th term of the AP will be 28.

Answer 2

$\large \underline{ \underline{ \sf \: Solution : \: \: \: }}$

Given ,

$\sf nth \: term = 3n-2$

So ,

$\sf A_{10}= 3 × 10 - 2 \\ \\ \sf</p><p>A_{10} = 30 - 2 \\ \\ \sf</p><p>A_{10} = 28$

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$\huge{ \sf \: Or \: }$

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Given ,

$\sf nth \: term \: (An) = 3n - 2 \\ \\ \sf</p><p>So \: , \: A_{1}= 3 × 1 - 2 = 1 \\ \\ \sf</p><p>A_{2}= 3 × 2 - 2 = 4 \\ \\ \sf</p><p>A_{3}= 3 × 3 - 2 = 7 \\ \\ \sf A_{4} = 3 × 4 - 2 = 10$

Hence , the first 4 terms of the sequence are 1 , 4 , 7 , 10 and first term = 4 , common difference = 3

We know that ,

$\large \fbox{ \fbox{ \sf \: nth \: term = a + (n - 1)d \: }}$

$\sf A_{10}= 1 + (10 - 1) × 3 \\ \\ \sf</p><p>A_{10} = 1 + 27 \\ \\ \sf A_{10} = </p><p>28$

Therefore , 28 is the 10th term of the given AP