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?
+2401 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. :)))
=^._.^=
+128407 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
