math prob

[Jennifer ^.^]

Sir, cud u pease tell me how to solve this????- conventionally and
otherwise...

>> in an infinite geometric progression, the sum of the first two terms is 6 and every term is four times the sum of all the terms that follow it. Find:
a) the geometric progression and
b) its sum upto infinity

i dint even understand the question btw....

[Arjun Mohan]

The series is 5,1,1/5...............

Sum will be 25/4

If this is given as answer options, its as simple as checking for the sum of 1st 2 terms and the ratio of 1st to rest of the answer options.

Conventionally

GP is a, ar, ar^2.....................

4ar/(1-r) = a [4 times the sum of terms starting from second term ar to infinity is the fist term 'a']

Solving this you get r = 1/5. Now create a GP whose r = 1/5 and first 2 terms sum as 6.

Regards,
Arjun

--
If ignorance is bliss,why do we seek knowledge?

[Jennifer ^.^]

oh...... ok- thanx!!!! :)

On Nov 22, 10:00 pm, Arjun Mohan <arjun.mo...@gmail.com> wrote:
> The series is 5,1,1/5...............
> Sum will be 25/4
> If this is given as answer options, its as simple as checking for the sum of
> 1st 2 terms and the ratio of 1st to rest of the answer options.
>
> Conventionally
> GP is a, ar, ar^2.....................
> 4ar/(1-r) = a [4 times the sum of terms starting from second term ar to
> infinity is the fist term 'a']
> Solving this you get r = 1/5. Now create a GP whose r = 1/5 and first 2
> terms sum as 6.
>
> Regards,
> Arjun
>

[Jennifer Durai]

HAPPY TEACHERS' DAY SIR!!!!!!!!!!!!!

By the way, this is your student from Byju's CET class :) Have a nice Day!!! :)

Truly,

           - Jennifer {aka Jeny}