Answer:
\(a_{30}=91\)
Step-by-step explanation:
Given : an arithmetic sequence 4, 7, 10,....
We have to find the 30th term of the given arithmetic series.
Consider the given arithmetic sequence 4, 7, 10,....
Here,
\(a_1=4\\\\ a_2=7\\\\ a_3= 10\)
We first find the common difference (d),
\(a_2-a_1=7-4=3\)
\(a_3-a_2=10-7=3\)
Since, the difference between the terms are same so the common difference is 3.
The formula for the general term in an arithmetic series is given by
\(a_n= a+(n-1)d\)
Where , n is the number i=of the term,
a is first term
d is common difference,
Since, we have to find the 30 th term,
So put n = 30 , d = 3 , a= 4
We have,
\(a_{30}=4+(30-1)3\)
Simplify, we have,
\(a_{30}=4+29\times 3\)
\(a_{30}=4+29\times 3\)
Thus, \(a_{30}=91\)
