Respuesta :

[tex]a_1=4;\ a_2=7;\ a_3=10\\\\d=a_2-a_1\to d=7-4=3\\\\a_n=a_1+(n-1)d\to a_n=4+(n-1)\cdot3=4+3n-3=3n+1\\\\\boxed{a_n=3n+1}\\\\a_{30}=3\cdot30+1=90+1=91\\\\Answer:\boxed{a_{30}=91}[/tex]

Answer:

[tex]a_{30}=91[/tex]

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,

[tex]a_1=4\\\\ a_2=7\\\\ a_3= 10[/tex]

We first find the common difference (d),

[tex]a_2-a_1=7-4=3[/tex]

[tex]a_3-a_2=10-7=3[/tex]

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

[tex]a_n= a+(n-1)d[/tex]

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,

[tex]a_{30}=4+(30-1)3[/tex]

Simplify, we have,

[tex]a_{30}=4+29\times 3[/tex]

[tex]a_{30}=4+29\times 3[/tex]

Thus, [tex]a_{30}=91[/tex]