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maksim [4K]
3 years ago
11

I need help with problems 14-15.

Mathematics
1 answer:
oee [108]3 years ago
7 0

For 14, the answer is >. You inversed it. You are right about question 15.


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2. Find the 20th term of an arithmetic sequence if its 6th term is 14 and 14th term is 6.
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Answer:

\sf t_{20}= 0

Step-by-step explanation:

<h3>Arithmetic sequence:</h3>

      \sf \boxed{\bf n^{th} \ term = a + (n-1)d}\\\\\text{Here, a is the first term ; d is the common difference }

6th term is 14 ⇒ \sf t_6 = 14

                a + (6 - 1)d = 14

                    a  +  5d = 14  --------------(I)

14th term is 6 ⇒\sf t_{14} = 6

             a + (14-1)d = 6

                  a + 13d = 6 ----------------(II)

Subtract equation (II) from equation(I)

        (I)          a + 5d = 14

        (II)         a + 13d = 6

                    <u>-    -          -</u>

                            -8d = 8

                               d  = 8 ÷(-8)      

                              \sf \boxed{\bf d= (-1)}

Plugin d = -1 in equation (I)

a + 5(-1) = 14

      a -5  = 14

             a = 14 + 5

             \sf \boxed{\bf a = 19}  

20th term:

 \sf t_{20}= 19 + 19*(-1)

       = 19 - 19

   \sf \boxed{\bf t_{20} = 0}

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Step-by-step explanation:

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