Question

Question #3f536

Answer 1

For your first, we'll write a system of equations since we have two variables

Explanation:

a is the first term and r the common ratio.
`t_n = a + (n - 1)d`

`1 = a + (3 - 1)d`

`1 - 2d = a`

Equation 2

`s_n = n/2(2a + (n - 1)d)`

I think it's easiest to add the first two terms on to the sum of the last five. We can do this with the expression `85 + a + (a + r)`

`85 + a + (a + d)= 7/2(2a + 6d)`

`85 + 2a + d= 7a + 21d`

Now, solve by substitution.

`85 + 2(1 - 2d) + d= 7(1 - 2d) + 21d`

`85 + 2 - 4d + d= 7 - 14d + 21d`

`80 = 10d`

`8 = d`

Therefore, `a = -15`.

To find the 6th term we use the formula `t_n = a + (n - 1)d`

`t_6 = -15 + (6 - 1)8`

`t_6 = -15 + 40`

`t_6 = 25`

To summarize, the common difference is 8, the first term is -15 and the sixth term is 25.

I'll leave the last problem for someone else to solve. If they don't, I'll come back and answer it for you.

Hopefully this helps!