2 years ago

If the 13th and 38th terms of an arithmetic sequence are -53 and -128,

Mathematics

1 answer:

love history [14]2 years ago

7 0

9514 1404 393

Answer:

-2018

Step-by-step explanation:

The n-th term is ...

an = a1 +d(n -1)

So, the given terms are ...

-53 = a1 +12d

-128 = a1 +37d

Subtracting the first from the second gives ...

(a1 +37d) -(a1 +12d) = (-128) -(-53)

25d = -75

d = -3

The 668th term will be ...

a668 = a1 +d(668 -1) = a1 +667d = (a1 +37d) +630d

a668 = -128 +630(-3) = -128 -1890 . . . . substitute for a38

a668 = -2018

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$\Large\textsf{\textbf{Answer\::}}$

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