Chapter ISG, Problem 7.10.1P

a.

To determine

To find recursive and explicit formula.

Answer to Problem 7.10.1P

The explicit formula is $a_n=3{(2)}^{n-1}$ and the recursive formula is $a_n=2a_{n-1}$ .

Explanation of Solution

Given information:

The geometric sequence is,

3,6,10

Formula Used:

Geometric sequence is,

  $a_n=a{(r)}^{n-1}$

Common ratio is,

  $r=\frac{a2}{a1}$

Calculation:

Common ratio is, substituting the values,

  $r=\frac{6}{3}$

On solving,

  $r=2$

For explicit formula, substituting in geometric sequence formula,

  $a_n=3{(2)}^{n-1}$

For recursive formula,

  $a_n=a_{n-1}\times 2$

Can rewrite it as,

  $a_n=2a_{n-1}$

Hence, the explicit formula is $a_n=3{(2)}^{n-1}$ and the recursive formula is $a_n=2a_{n-1}$ .

Conclusion:

The explicit formula is $a_n=3{(2)}^{n-1}$ and the recursive formula is $a_n=2a_{n-1}$ .

b.

To determine

To find recursive and explicit formula.

Answer to Problem 7.10.1P

The explicit formula is $a_n=9n-12$ and the recursive formula is $a_n=a_{n-1}+9$ .

Explanation of Solution

Given information:

The arithmetic sequence is,

-3, 6, 15

Formula Used:

Arithmetic sequence is,

  $a_n=a+d(n-1)$

Common ratio is,

  $d=a2-a1$

Calculation:

Common difference is, substituting the values,

  $r=6-(-3)$

So,

  $r=6+3$

On solving,

  $r=9$

For explicit formula, substituting in sequence formula,

  $a_n=-3+9(n-1)$

So,

  $a_n=-3+9n-9$

On solving,

  $a_n=9n-12$

For recursive formula, substituting the values

  $a_n=a_{n-1}+9$

Hence, the explicit formula is $a_n=9n-12$ and the recursive formula is $a_n=a_{n-1}+9$ .

Conclusion:

The explicit formula is $a_n=9n-12$ and the recursive formula is $a_n=a_{n-1}+9$ .