Question

30th term of the AP: 1,7,13........is:

Answer 1

Answer:

a = 1

d = 7 - 1 = 6

n = 30

$an \: = a \: + (n - 1)d \\ \\ a30 = a \: + (30 - 1)d \\ \\ = 1 + 29 \times 6 \\ \\ = 1 + 174 \\ \\ = 175$

So the 30th term is 175.

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Answer 2

The 30th term of the AP 1 , 7 , 13 , . . . is 175

Given :

The arithmetic progression 1 , 7 , 13 , . . .

To find :

The 30th term of the AP

Concept :

If in an arithmetic progression

First term = a

Common difference = d

Then nth term of the AP

= a + ( n - 1 )d

Solution :

Step 1 of 3 :

Write down the given progression

Here the given arithmetic progression is

1 , 7 , 13 , . . .

Step 2 of 3 :

Write down first term and common difference

The arithmetic progression is

1 , 7 , 13 , . . .

First term = a = 1

Common Difference = d = 7 - 1 = 6

Step 3 of 3 :

Find the 30th term of the AP

Then 30th term of the AP

= a + ( 30 - 1 )d

= a + 29d

= 1 + ( 29 × 6 )

= 1 + 174

= 175

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