Question

Find the nth term of each of these sequences
a)16,19,22,25,28

b)1,3,9,27,81

Answer 1

a) 16+(n-1)3=n th term. b)
${3}^{n - 1}$

Answer 2

The $n^{th}$ term of the given sequences are a)13 + 3n and b) $3^{n -1}$.

Given:

The sequences a)16,19,22,25,28 and b)1,3,9,27,81

To Find:

The $n^{th}$ term of the given sequences.

Solution:

a) A sequence is said to be in arithmetic progression if the difference between the consecutive terms is always constant.

Let us analyze the given sequence 16, 19, 22, 25, 28.

Let us find the difference between each of the consecutive terms.

19 - 16 = 3, 22 - 19 = 3, 25 - 22 = 3, 28 - 25 = 3.

We observe that the difference between each of the consecutive terms is 3. Hence the sequence is in arithmetic progression with

The common difference, d = 3

The first term, a = 16.

The $n^{th}$ term of an AP, $a_{n}$  = a + (n - 1)d = 16 + (n - 1)3

⇒ $a_{n}$  = 16 + 3n - 3 = 13 + 3n.

∴ The $n^{th}$ term of the given sequence is 13 + 3n.

b)  A sequence is said to be in geometric progression if the ratio between every term to its immediate predecessor in the sequence is always the same (and non-zero).

Let us analyze the given sequence 1, 3, 9, 27, 81.

Let us find the ratio of each of the consecutive terms.

3/1 = 3, 9/3 = 3, 27/9 = 3, 81/27 = 3

We observe that the ratio of each of the terms to its previous term in the sequence is 3. Hence the sequence is in geometric progression with:

The common ratio, r = 3.

The first term, a = 1.

The $n^{th}$ term of an G, $T_{n}$  = a$r^{n -1}$.

⇒   $T_{n}$  = 1 x $3^{n -1}$ =  $3^{n -1}$

∴ The $n^{th}$ term of the given sequence is $3^{n -1}$ .

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