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faust18 [17]
3 years ago
8

What is the twelfth term in the sequence 13, 26, 39, …?

Mathematics
2 answers:
OverLord2011 [107]3 years ago
8 0

Answer:

the answer is 156

Step-by-step explanation:

for my own method

13 is the sequence

so 13×12=156

<h3> or</h3>

13+13=26+13=39+13=52+13=65+13=78+13=91+13=104+13=117+13=130+13=143+13=156.

<h2><em>s</em><em>o</em><em> </em><em>t</em><em>h</em><em>a</em><em>t</em><em> </em><em>i</em><em>s</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>f</em><em>i</em><em>n</em><em>a</em><em>l</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em></h2>
Gwar [14]3 years ago
7 0

Answer:

  • 156

Step-by-step explanation:

<u>Given the sequence </u>

  • 13, 26, 39, …

<u>We see it is an AP with</u>

  • The first term a = 13
  • Common difference d = 13 ( as 26 - 13 = 39 - 26 = 13)

<u>The 12th term is:</u>

  • a₁₂ = a + 11d = 13 + 11*13 = 156
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