Given `3`, `m`, `n`, `192` are the first four consecutive terms of a geometric progression. Find the three consecutive terms which added up to 16128 ?

In a geometric progression if firs term is `a_1` and common ratio is `r`, subsequent tems are `a_2=a_1r`, `a_3=a_1r^2` and `a_4=a_1r^3` i.e. `n^(th)` term is `a_n=a_1r^(n-1)`

In the first four consecutive terms of geometric progression, we have `3,m,n,192` i.e. `a_1=3` and `a_1r^3=192`

i.e. `3r^3=192` or `r^3=192/3=64` and hence `r=root(3)64=4`

and then `m=3*4=12` and `n=12*4=48`

Let the three consective terms which add up to `16128` be `a_u,a_(u+1)` and `a_(u+2)`

and `a_u=3*4^(u-1)`, `a_(u+1)=3*4^u` and `a_(u+2)=3*4^(u+1)` and their sum is

`3(4^(u-1)+4^u+4^(u+1))=16128`

or `4^(u-1)+4^u+4^(u+1)=5376`

or `4^(u-1)(1+4+16)=5376`

or `4^(u-1_=5376/21=256`

or `4^(u-1)=4^4`

i.e. `u-1=4` or `u=5`

Hence three consective terms whose sum is `16128` are `3*4^4, 3*4^5` and `3*4^6` i.e. `768,3072` and `12288`

The nth term of a geometric progression is given by:

`bb(ar^(n-1))`

Where:

`bba` is the first term, `bbr` is the common ratio, `bbn` is the nth term.

We know if `a,b,c` are in geometric sequence, then:

`b/a=c/b`

Using this we can find the value of `m`

`3(m/3)^(0)`

`3(m/3)^(3)=192`

Dividing by 3:

`(m/3)^3=64`

Taking cubed root:

`m/3=4`

`m=12`

We know if `a,b,c` are in geometric sequence, then:

`b/a=c/b`

`:.`

`m/3=192/n`

Substituting `m=12`

`12/3=192/n`

`4=192/n`

`n=192/4=48`

We can now find the common ratio:

`12/3=192/48=4`

The sum of a geometric series is given by:

`a((1-r^(n))/(1-r))`

Where:

`bba` is the first term, `bbr` is the common ratio, `bbn` is the nth term.

We know the sum we require is `16128`, and we know the common ratio. `4`, and we need 3 consecutive terms.

We need to find `a`:

`a((1-(4)^3)/(1-4))=16128`

`a((-63)/(-3))=16128`

`a(21)=16128`

`a=16128/21=768`

So our 3 consecutive terms are:

`768(4)^0 , 768(4)^1,768(4)^2`

`color(blue)(768, 3072,12288)`