I’m not sure how to solve and get the answer i need with both

Step-by-step explanation:

1) we want to find the 12th term of the arithmetic sequence given that the 7th term is 40 and the 18th term is 106

Recall

Tn = a + (n-1)d

Where Tn is the value of the nth term

a is the value of the first term

d is the common difference between a term and the following term

n is the number of terms

For the 7th term,

40 = a + (7-1)d

40 = a + 6d - - - - - - -1

For the 18th term,

106 = a + (18-1)d

106 = a + 17d - - - - - - -2

Subtract equation 2 from equation 1

-66 = -11d

d = 66/11 = 6

Put d = 6 in equation 1

a = 40-6d

a = 40 - 6×6

a = 40 - 36 = 4

To find the 7th term,

T 12 = 4 + (12-1)6

T12 = 4 + 11×6 =4 +66 =70

2) we want to find the 7th term of the geometric sequence given that the 2nd term is 324 and the 4th term is 36. Recall

Tn = ar^(n-1) where

a is the value of the first term

r is the common ratio between a term and the previous term

n is the number of terms.

For 2nd term

324 = ar^(2-1)

324 = ar- - - - - - -1

For 4th term

36 = ar^(4-1)

36 = ar^3 - - - - - -2

Divide equation 1 by equation 2

324/36 = ar / ar^3

9 = r^(1-3) = r^-2

9 = 1/r^2

9r^2 = 1

r^2 = 1/9

Take square root of both sides

r = 1/3

Put r = 1/3 in equation 1

324 = 1/3 × a

a/3 = 324

a = 3 × 324 = 972

To find the 7th term,

T7 = 972 × 1/3^(7-1)

T7 = 972 × (1/3)^6

T7 = 972 × 0.00137174211

T7 = 4/3

7th term is 4/3

the reason why your question keeps getting deleted because this is not a school related question : )

i will answer your question though.

i never tried church's

i dont know

step-by-step explanation: