Help my question pls! Arithmetic sequence

Question

In an arithmetic sequence, the 9th term is twice the 3rd term and the 15th term is 80. Find the common difference and the sum of the terms from the 9th to the 15th inclusive.

Answer 1

Let X = 3rd term,, Y = 2X = 9th term and 80 = 15th term

Now Y= X + 6d, where d is the difference between each term and 80 = Y + 6d

Subtract leftmost equation from rightmost equation in second line to eliminate d getting

80 - Y = Y - X or 80 = 2Y - X

But from line 1 Y = 2X

Substitue for Y getting 80 = 4X - X = 3X

Solving we get X = 80/3 and we see that Y = 2X = 160/3

To get d we yse 80 = Y + 6d. or 80 = 160/3 + 6d and solving we get d = 80/18.

As a check we see if we get 80 when we add 6d to 160/3

So, 160/3 + 6 (80/18) or (160 + 80) / 3 = 240 / 3 = 80 as it should.

I leave the sum to you. The fact that the term are not integers is not a problem, as nothing in the problem statement said they had to be.

Answer 2

a₉ = 2a₃

a₉ = a₃ + 6d

2a₃ = a₃ + 6d

a₃ = 6d

a₃ = a₁ + 2d

6d = a₁ + 2d

4d = a₁

80 = a₁ + 14d

80 = 4d + 14d

80 = 18d

5 = d

80 = a₁ + 14d

80 = a₁ + 80

0 = a₁

a₉ = 0 + 40

a₉ = 40

a₁₅ = 70

S = 3(40 + 70)

S = 3(110)

S = 330

80