Find the 19th term for the arithmetic sequence in which the 23rd term is 9 on the 11th term is -75

an = a1+d(n-1)

9 = a1+d(23-1)

9 = a1 +22d

-75 = a1+ d(11-1)

-75 = a1 +10d

subtract the 2 equations

9 = a1 +22d

-75 = a1 +10d

9--75 = 22d-10d

84 = 12d

divide by 12 on each side

84/12 = 12d/12

7 = d

now substitute in here to find a1

9 = a1 +22d

9 = a1 + 22(7)

9 = a1 + 154

subtract 154 from each side

-145 = a1

an =-145 + 7(n-1)

a19 = -145 + 7(19-1)

a19 = -145 + 7*18

a19 = -19

1) c 2143.6 cm

2) c 113.0 m

3) a 33.5 ft

4)   d 2 cm

5) c 20 mm

6) 180 in

7) d 120 ft

true

step-by-step explanation: