contestada

A sequence has a common second difference of 4 between terms. The first two terms are 3 and 12. Write down the 3rd and 4th terms.​

Respuesta :

onyebuchinnaji

The first two terms of the sequence are 3 and 12 and the 3rd and 4th terms are 25 and 42.

3rd and 4th terms from second level difference

The 3rd and 4th terms of the sequence is determined from second level difference as shown below;

  • let the first term = a
  • second term = b
  • third term = c
  • fourth term = d

(c - b) - (b - a) = 4  ----(1)

(d - c) - (c - b) = 4 ---- (2)

from (1);

(c - 12) - (12 - 3) = 4

(c - 12) - (9) = 4

c - 21 = 4

c = 25

from (2);

(d - 25) - (25 - 12) = 4

d - 38 = 4

d = 42

The sequence = 3, 12, 25, 42

check

first difference = (12 - 3), (25-12), (42-25) = 9, 13, 17

second difference = (13 - 9), (17 - 13) = 4, 4

Thus, the first two terms of the sequence are 3 and 12 and the 3rd and 4th terms are 25 and 42.

Learn more about sequence here: https://brainly.com/question/6561461

#SPJ1

Otras preguntas