Question

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.

Answer 1

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

Answer 2

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