Question

Identify the 35th term of an arithmetic sequence where a1 = −7 and a18 = 95. (2 points)

Answer 1

To answer the problem, determine first the common difference by,
 
                                     d = (a18 - a1) / (18 - 1)
 
                                     d = ( 95 - -7) / (18 - 1) = 6

To solve for the 35th term
 
                                      an = a1 + (n - 1)d
      
                                     a35 = -7 + (35 - 1)6 = 197

Therefore, the 35th term is 197. 

Answer 2

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.