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