Question

Write an equation for the nth term of the arithmetic sequence.

Then find a50.

3, 9, 15, 21, . . .

Answer 1

`\bold{a_n=6n-3}\\\\\bold{a_(50)=297}`

Explanation:

3, 9, 15, 21, ...

9-3 = 6; 15-9 = 6; 21-15 = 6 ⇒ d = 6

first term: a = 3

difference: d = 6

so the formula:

`a_n=a+d(n-1)\\\\a_n=69+6(n-1)\\\\a_n =3+6n-6\\\\\underline{a_n=6n-3}`

and:

`a_(50)=6\cdot50-3=300-3=297`