Find the first 5 terms of the arithemitic sequence whose seventh term is 22 and whose eleventh term is 6.

The first 5 terms of the arithemitic sequence are 46, 42, 38, 34 and 30.

Step-by-step explanation:

Given that:

The arithemitic sequence whose seventh term is 22 and whose eleventh term is 6.

To Find:

The first 5 terms of the arithemitic sequence.

We know that:

aₙ = a + (n - 1)d

Where,

aₙ = nth term a = First term n = Number of termsd = Common difference

We have:

Seventh term = 22

⟶ a + (7 - 1)d = 22

⟶ a + 6d = 22

⟶ a = 22 - 6d

Evelenth term = 6

⟶ a + (11 - 1)d = 6

⟶ a + 10d = 6

Substituting the value of a.

⟶ 22 - 6d + 10d = 6

⟶ 22 + 4d = 6

⟶ 4d = 6 - 22

⟶ 4d = - 16

⟶ d = - 16/4

⟶ d = - 4

In equation (i).

⟶ a = 22 - 6d

Putting the value of d.

⟶ a = 22 - 6(- 4)

⟶ a = 22 + 24

⟶ a = 46

Now we have to find the first 5 terms:

a = 46a + d = 46 + (- 4) = 46 - 4 = 42a + 2d = 46 + 2(- 4) = 46 - 8 = 38a + 3d = 46 + 3(- 4) = 46 - 12 = 34a + 4d = 46 + 4(- 4) = 46 - 16 = 30

Hence,

The first 5 terms of the arithemitic sequence are 46, 42, 38, 34 and 30.

Y=-2x-6 is the answer