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find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule A(n)=-6+(n-1)(1/5)

Respuesta :

sqdancefan
A(1) = -6 +(1 -1)*(1/5) = -6

A(4) = -6 +(4 -1)*(1/5) = -5 2/5

A(10) = -6 +(10 -1)*(1/5) = -4 1/5

Answer with explanation:

The formula that represents the nth term of a arithmetic sequence is given by:

[tex]A(n)=-6+(n-1)(\dfrac{1}{5})[/tex]

Now, we are asked to find the  first, fourth, and tenth terms of the arithmetic sequence.

i.e. we are asked to find the value of A(n) when n=1 ,4 and 10

  • when n=1 we have:

[tex]A(1)=-6+(1-1)(\dfrac{1}{5})\\\\i.e.\\\\A(1)=-6+0\\\\i.e.\\\\A(1)=-6[/tex]

  • now when n=4 we have:

[tex]A(4)=-6+(4-1)\times (\dfrac{1}{5})\\\\i.e.\\\\A(4)=-6+3\times \dfrac{1}{5}\\\\i.e.\\\\A(4)=-6+\dfrac{3}{5}\\\\i.e.\\\\A(4)=\dfrac{-6\times 5+3}{5}\\\\i.e.\\\\A(4)=\dfrac{-30+3}{5}\\\\i.e.\\\\A(4)=\dfrac{-27}{5}[/tex]

  • when n=10 we have:

[tex]A(10)=-6+(10-1)\times (\dfrac{1}{5})\\\\i.e.\\\\A(10)=-6+9\times \dfrac{1}{5}\\\\i.e.\\\\A(10)=-6+\dfrac{9}{5}\\\\i.e.\\\\A(10)=\dfrac{-6\times 5+9}{5}\\\\i.e.\\\\A(10)=\dfrac{-30+9}{5}\\\\i.e.\\\\A(10)=\dfrac{-21}{5}[/tex]

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