Question

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

A(n) = –6 + (n–1)(6)


6, 18, 54


–6, 12, 48


0, 18, 54


–6, 18, 54

Answer 1

Simplify:

A(n) = -6 + (n - 1)(6) = -6 + (n)(6) + (-1)(6) = -6 + 6n - 6 = 6n - 12

Put n = 1, n = 4 and n = 10 to the expression:

A(1) = 6(1) - 12 = 6 - 12 = -6

A(4) = 6(4) - 12 = 24 - 12 = 12

A(10) = 6(10) - 12 = 60 - 12 = 48

Answer:

first term = -6, fourth term = 12 and tenth term = 48.

-6, 12, 48

Answer 2

-6,12 and 48.

Explanation:

A(n) = –6 + (n–1)(6)

This is the the nth term.

The first term is obtained when we use n = 1:-

First term A(1) = -6 + (1 -1)(6) = -6 + 0

= -6.

To find the fourth term we replace n by 4 :

A(4) = -6 + (4 - 1)(6)

= -6 + 18

= 12.

In a similar way the 10th term

= -6 + (10-1)(6)

= -6 + 54

= 48 .