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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.

 Mar 6, 2020
 #1
avatar+21874 
+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.

 Mar 6, 2020
 #3
avatar+111435 
+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

 

 

 

cool cool cool

 Mar 7, 2020

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