Question

25PTS!!

Answer 1

Answer:

The recursive formula is given as:

$a_n = 3a_{n-1}+2$ , n is the number of terms

where,

$a_4 = 20$

We have to find  first three terms of the sequence.

For n =4 ;

$a_4 = 3a_{4-1}+2$

⇒$a_4 = 3a_3+2$

⇒$20 = 3a_3+2$

Subtract 2 from both sides we have;

$18= 3a_3$

Divide both sides by 3 we have;

$6 = a_3$

or

$a_3 =6$

For n = 3;

$a_ 3= 3a_{2}+2$

⇒$6 = 3a_2+2$

Subtract 2 from both sides we have;

$4= 3a_2$

Divide both sides by 3 we have;

$a_2 = \frac{4}{3}$

For n =2

$a_2= 3a_{1}+2$

⇒$\frac{4}{3} =3a_1+2$

Subtract 2 from both sides we have;

$-\frac{2}{3}= 3a_3$

Divide both sides by 3 we have;

$a_1 =-\frac{2}{9}$

Therefore, the first three terms of the sequence  are:

$a_1 =-\frac{2}{9}$

$a_2 = \frac{4}{3}$

$a_3 =6$

Answer 2

The answer is
a1 =-2/9
a2 =4/3
a3 =6