Question

which term of the series 3 8 13 18 is 98

Answer 1

Hello!
Here is your answer

a ( first term) = 3

common difference (d) = 8-3 = 5

last term (T) = 98

number of terms ( n )= ?

By using the formula

T = a + ( n - 1 ) d

98 = 3 + ( n - 1 ) 5

95 = 5n - 5

100 = 5n

n = 20.....

Answer is 20

Answer 2

98 is 20th term of series 3 , 8 , 13 , 18 , . . . . .

Given :

The series 3 , 8 , 13 , 18 , . . . . .

To find :

which term of the series 3 , 8 , 13 , 18 , . . . . . is 98

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 series

Here the given series is

3 , 8 , 13 , 18 , . . . . .

This is an arithmetic series

Step 2 of 3 :

Write down first term and common difference

First term = a = 3

Common Difference = d = 8 - 3 = 5

Step 3 of 3 :

Find which term of the series is 98

Let nth term of the AP = 98

a + ( n - 1 )d = 98

⇒ 3 + ( n - 1 ) × 5 = 98

⇒ 3 + 5n - 5 = 98

⇒ 5n - 2 = 98

⇒ 5n = 100

⇒ n = 100/5

⇒ n = 20

Hence 98 is 20th term of series 3 , 8 , 13 , 18 , . . . . .

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