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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! :)