Answer:
Option (b) is correct.
35th term of an arithmetic sequence is 197.
Step-by-step explanation:
Given : \(a_1 =-7\ and\ a_{18}=95\)
We have to find the 35 term of an arithmetic sequence whose \(a_1 =-7 \ and\ a_{18}=95\)
For an Arithmetic sequence the general term is given as \(a_n=a+(n-1)d\),
where a is first term ,
n is number of term
d is common difference,
Thus, the 18th term \(a_{18}=a+(18-1)d\)
We have \(a_{18}=95\) , solving for d , we have,
\(a_{18}=a+(18-1)d=95\)
\(95=a+17d\)
Thus, \(\Rightarrow 95=-7+17d\)
\(\Rightarrow 95+7=17d\)
\(\Rightarrow 102=17d\)
Divide both side by 17, we get,
\(\Rightarrow d=6\)
Now, 35th term of an arithmetic sequence is given by,
\(\Rightarrow a_{35}=a+(35-1)d\)
Substitute, a and d , we get,
\(\Rightarrow a_{35}=-7+34(6)\)
\(\Rightarrow a_{35}=-7+204\)
\(\Rightarrow a_{35}=197\)
Thus, 35th term of an arithmetic sequence is 197.
Option (b) is correct.
