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We have four numbers written in a row. The first three numbers are in AP (arithmetic progression) of common difference equal to 6 and the last three numbers are in GP (geometric progression) of unknown common ratio. The first number is equal to the fourth number.

What is the sum of all of these numbers?

 Jan 27, 2021
 #1
avatar+382 
+1

Our numbers are a, a + 6, a + 12, a

Since the ratios are the same...

a/(a+12) = (a + 12)/ (a + 6)

a^2 + 6a = a^2 + 24a + 144

a = -8

So our sequence is -8, -2, 4, -8. 

-8 - 2 + 4 - 8 = -14. 

So our answer is -14. 

 

 I hope this helped. :)))

 

=^._.^=

 Jan 27, 2021
 #2
avatar+116126 
+1

Let N  be  the first number

 

So  the first three numbers  are   ( N ) + (N + 6) +( N + 12) 

 

The  last  three  numbers  are   (  N + 6 ) , (N + 6)r  + (N + 6)r^2

 

So

AP third number = GP second number

N +  12   = ( N + 6)r

(N + 12/(N +6)   = r

 

And  

AP first number = GP last number

N  = (N + 6)r^2

N = ( N + 6)   (N +12)^2  /(N + 6)^2

N = ( N + 12)^2  / (N + 6)

N (N + 6)  = (N + 12)^2

N ^2  + 6N  =N^2  + 24N + 144

-18N = 144

N = -8

r =  (N + 12) / (N + 6)  =  ( 4/-2)  = -2

 

The series  is    -8 , -2  , 4 ,  -8

 

The sum is   -14

 

 

cool cool cool

 Jan 27, 2021

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