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The 24th term of the arithmetic sequence where a1 = 8 and a9 = 56 is 146

Arithmetic sequence

  • First term, a1 = 8
  • Ninth term, a9 = 56
  • Common difference = d

a9 = a + 8d

56 = 8 + 8d

56 - 8 = 8d

48 = 8d

divide both sides by 8

d = 48 / 8

d = 6

24th term = a + 23d

= 8 + 23(6)

= 8 + 138

= 146

Therefore, the 24th term of the arithmetic sequence where a1 = 8 and a9 = 56 is 146

Learn more about arithmetic sequence:

https://brainly.com/question/6561461

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