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

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

7,24

step-by-step explanation: