Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.A(n) = 1 + (n – 1)(–5.7)


1, –21.8, –56


–5.7, –21.8, –51.3


1, –16.1, –50.3


0, –17.1, –51.3


If someone answers this, i'd like if someone could help me to know how to solve it too, please and thank you!

Answers

Answer 1
Answer:

1, - 16.1, - 50.3 ← third on list

To find the required terms, substitute n = 1, 4, 10 into the rule and evaluate

A(1) = 1 + (1 - 1)(- 5.7) = 1 + 0 = 1

A(4) = 1 + (4 - 1)(- 5.7 ) = 1 + (3 × - 5.7 ) = 1 - 17.1 = - 16.1

A(10) = 1 + (10 - 1 )(- 5.7) = 1 + (9 × - 5.7) = 1 - 51.3 = - 50.3


Answer 2
Answer:

\bf \stackrel{A(n)=1+(n-1)(-5.7)}{ \begin{array}{ll} n\qquad \qquad &A(n)\n \cline{1-2} 1&1+(1-1)(-5.7)\n &1+0\n &1\n\n 4&1+(4-1)(-5.7)\n &1+(3)(-5.7)\n &1-17.1\n &-16.1\n\n 10&1+(10-1)(-5.7)\n &1+(9)(-5.7)\n &1-51.3\n &-50.3 \end{array} }


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Answers

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