Find the 31st term of an arithmetic sequence where the eighth term is 45 and the fourth term is 65.

Let a(n) denote the nth term in the sequence. Then

a(5) = a(4) + d

a(6) = a(5) + d = a(4) + 2d

a(7) = a(6) + d = a(4) + 3d

a(8) = a(7) + d = a(4) + 4d

where d is the fixed difference between consecutive terms in the sequence.

We have a(4) = 65 and a(8) = 45, so we solve for d:

45 = 65 + 4d  ==>  d = -5

so, given some term in the sequence a(n), the next term is a(n + 1) = a(n) - 5.

Continuing the pattern above, we would wind up getting

a(31) = a(4) + 27d

(In case the pattern isn't clear, watch the coefficient of d; then 1 + 4 = 5, 2 + 4 = 6, 3 + 4 = 7, and so on.)

Then the 31st term in the sequence must be

a(31) = 65 + 31(-5) = -90

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

step-by-step explanation:

no, her answer is not correct.   we square the lengths of all sides and then determine whether the pythagorean theorem holds here:

9² + 15² = 306.   this is not equal to the square of the longest side ( that square being 395.   answer c is correct.