Whats the formula of the sequence 13, 27, 41, 55, ?
A. a^n = 13 + 14(n-1) B. a^n = 13(n-1) C. a^n = 13(14)^(n-1)
D. a^n = 14 + 13(n-1)

A

Step-by-step explanation:

Note the common difference between consecutive terms in the sequence.

d = 27 - 13 = 41 - 27 = 55 - 41 = 14

This indicates the sequence is arithmetic with n th term

= a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 13 and d = 14, thus

= 13 + 14(n - 1) → A

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240