The first, third and ninth terms of a arithmetic sequence (A.P) are equal to the first three terms of a geometric sequence (G.P). if the seventh term of the linear sequence is 14.

Calculate:

(a) the twentieth term of the arithmetic sequence

(b) the sum of the first twelve term of the G.P.

Guest Mar 6, 2020

#1**+1 **

Let the first term of the arithmetic sequence = a

and the common difference of the arithmetic sequence = d.

Let the first term of the geometric sequence = a

and the common ratio of the geometric sequence = r.

By some quick guessing, I came up with these values: a = 2 d = 2 r = 3.

From these, you can calculate the vaues you desire.

geno3141 Mar 6, 2020

#3**+2 **

Let the first term of the arithmetic series = a

Let the third term = a + 2d

Let the ninth term = a + 8d

And we know that a + 6d = 14

The second term of the geometric series = ar

The thrid term of the geometric series = ar^2

So

a + 2d = ar ⇒

2d = ar - a ⇒

2d = a ( r - 1)

a + 8d = ar^2 ⇒

8d = ar^2 - a ⇒

8d = a (r^2 - 1) ⇒

8d = a (r- 1) (r + 1) ⇒

8d = (2d)(r + 1) ⇒

8 = 2 (r + 1) ⇒

4 = r + 1

So r = 3

And

2d = a (3 - 1)

2d = 2a

d = a

So

a + 6d = 14

a + 6a = 14

7a = 14

a = 2 so d also = 2

(a) The twelveth term of the arithmetic series = a + 11d = 2 + 11(2) = 24

(b) The sum of the first 12 terms of the geometric series is

S = a [ 1 - r^12 ] [ 1 - r ] = 2 [ 1 - 3^12] / [ 1 - 3 ] = [ 3^12 - 1 ] = 531440

CPhill Mar 7, 2020