Question

Identify the 12th term of the arithmetic sequence in which a7 = 40 and a18 = 106.

Answer 1

we are given

airthematic sequence

we can use nth term formula

$a_n=a_1+(n-1)d$

n is number of terms

an is nth term

d is common difference

a1 is first term

We are given

a7=40

we can use it

$a_7=a_1+(7-1)d$

$40=a_1+6d$

$a_1+6d=40$

a18 = 106

$a_1_8=a_1+(18-1)d$

$a_1+17d=106$

we can subtract both equations

$a_1+17d-a_1-6d=106-40$

$11d=66$

$d=6$

now, we can find a1

$a_1+6d=40$

we can plug d=6

$a_1+6*6=40$

$a_1=4$

12th term:

$a_1_2=a_1+(12-1)d$

now, we can plug values

$a_1_2=4+(12-1)*6$

$a_1_2=4+66$

$a_1_2=70$

so, 12th term is 70...........Answer

Answer 2

The 12th term of the given arithmetic sequence is; a₁₂ = 70

How to find the nth term of an Arithmetic Sequence?

The formula for nth term of an arithmetic sequence is;

aₙ = a + (n - 1)d

where;

a is first term

n is position of term

d is difference between consecutive terms

We are given;

a₇ = 40

a₁₈ = 106

Thus;

a + (7 - 1)d = 40

a + 6d = 40    ------(1)

a + 17d = 106   ------(2)

Subtract eq 1 from eq 2 to get;

11d = 66

d = 6

Thus;

a + 6(6) = 40

a = 4

Thus;

a₁₂ = 4 + (12 - 1)6

a₁₂ = 70

Read more about Arithmetic Sequence at; https://brainly.com/question/6561461

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