3 years ago

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

-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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