Question

A sequence has an initial value (I) of -17 and its fourth term (a4) is 35. What is its general equation? Its 14th term?

Answer 1

Answer:

1. The sequence is an Arthemtic Progression

An=A1+(n-1)d

A1 is first term, An is nth term, n is number of term, and d is common difference

therefore

A4=35, A1= -17

A4=A1+(4-1)d

35= -17+3d

35+17=3d

52=3d

52/3=3d/3

14=d

common diffrence(d)=14

• The general solution is given by

An= -17+(n-1)14

An= -17+14n-14

An= -31+14n

An= 14n-31

A14 term, means n=14

From An=A1+(n-1)d

A14= -17+(14-1)14

= -17+(13×14)

= -17+182

= 165.

Therfore, the 14th term is 165.

2. A sequence has a CR of 4/5 and its eighth term (a8) is (393216/3125). What is its general equation? Its 3rd term?

solution

common ratio(r)=4/5

eighth term(G8)=393216/3125

From Gn= G1r^(n-1)

G8 means n=8

G8=G1r^(n-1)

393216/3125=G1(4/5)^(8-1)

393216/3125=G1(4/5)^7

G1=(393216/3125)/(4/5)^7

G1=600

The first term is given by G1=600

• Therefore

The General equation is given by

The General equation is given by Gn= 600(4/5)^(n-1)

3rd term (G3)

G3= G1(4/5)^(3-1) where n=3,

=600(4/5)^2

=600(16/25)

=384

Therefore, the 3rd term is given by G3= 384.

I have made some Corrections i messed up somewhere.