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

i dont get it. where are the "descriptions"?

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.