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Gopika
Sep 13, 2010, 02:10 PM
The fourth, seventh and sixteenth terms of an Arithmetic Progression are in a geometric progression, If the first six terms of the Arithmetic progression have a sum of 12, find the common difference of the Arithmetic Progression and the common ratio of the Geometric Progression
Unknown008
Sep 14, 2010, 07:49 AM
Ok, what is given:
T_4, T_7, T_{16}
are in geometric progression.
So, we know that:
\frac{T_7}{T_4} = \frac{T_{16}}{T_7}
This can be written as:
\frac{a + (7-1)d}{a + (4-1)d} = \frac{a + (16-1)d}{a + (7-1)d}
Then, we are given:
S_6 = \frac{6}{2}(2a + (6-1)d) = 12
You have two equations with two unknowns. I think you can now find the answer. Post what you get! :)
Gopika
Sep 14, 2010, 12:36 PM
I got two and then three
Thank you very much Unknown 008.. its helped me
Unknown008
Sep 14, 2010, 09:44 PM
Well done! :)