Question
Find the position of 98 in the following series 3,8,13 ....?
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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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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More Questions on This Topic :
Question 9. In an A.P T4:T7::2:3 then find T3:T11? ....
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
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
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