Question #3f536

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Answer 1 · 2016-03-21T16:02:04

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

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!