Find the 30h term of the geometric sequence whose initial term is 0 and the common ration is 4​

Sagot :

Answer:

Answer:

1. a6 = 33,614

2. a7 = 46,875

3. a 30 =0

4. a16 = -98,304

5. a8 = -559,872

Step-by-step explanation:

We will use the equation a_{n} =a_{1} r^{n-1}a

n

=a

1

r

n−1

1. a1 = 2; r = 7; 6th term

a_{6} =2( 7^{6-1})a

6

=2(7

6−1

)

a_{6} =2( 7^5)a

6

=2(7

5

)

a_{6} =2( 16807)a

6

=2(16807)

a6 = 33,614

2. a1 = 3; r = 5; 7th term

a_{7} =3( 5^{7-1})a

7

=3(5

7−1

)

a_{7} =3( 5^{6})a

7

=3(5

6

)

a_{7} =3( 15625)a

7

=3(15625)

a7 = 46,875

3. a1 = 0; r = 4; 30th term

a_{30} =0( 5^{30-1})a

30

=0(5

30−1

)

a_{30} =0( 5^{29})a

30

=0(5

29

)

a_{30} =(0)(86,264,514,923,095,703,125)a

30

=(0)(86,264,514,923,095,703,125)

a 30 =0

4. a1 = -3; r = 2; 16th term

a_{16} =-3( 2^{16-1})a

16

=−3(2

16−1

)

a_{16} =-3( 2^{15})a

16

=−3(2

15

)

a_{16} =-3( 32786)a

16

=−3(32786)

a16 = -98,304

5. a1 = -2; r = 6; 8th term

a_{8} =-2( 6^{8-1})a

8

=−2(6

8−1

)

a_{8} =-2( 6^{7})a

8

=−2(6

7

)

a_{8} =-2( 279936)a

8

=−2(279936)

a8 = -559,872