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Question

Find the position of 98 in the following series 3,8,13 ....?

Options:
A .  12th term
B .  20th term
C .  34th term
D .  36th term
E .  None of these
Answer: Option B
To find the position of 98 in the given series 3, 8, 13, ..., we need to first determine the pattern of the series.
The given series is an arithmetic sequence where the common difference is 5. This means that each term in the sequence is obtained by adding 5 to the previous term.
To find the position of 98 in the sequence, we can use the following formula to find the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
where:an = the nth term of the sequencea1 = the first term of the sequenced = the common differencen = the position of the term we want to find
We know that a1 = 3 and d = 5, and we want to find the value of n for which an = 98. So we can rearrange the formula as follows:
n = (an - a1)/d + 1
Substituting the given values, we get:
n = (98 - 3)/5 + 1n = 20
Therefore, the position of 98 in the given series is the 20th term. So the correct option is B.

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4 Comments

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
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
a=3,d=5,tn=98
formula,tn=a+(n-1)d
98=3+(n-1)5
98=3+5 n-5
98=-2+5 n
98+2=5 n
100=5 n
n=20
ans 20th term
Solution

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