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 ?

2 Answers
Shwetank Mauria
Apr 16, 2018

Three consective terms whose sum is #16128# are #768,3072# and #12288#

Explanation:

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#

Somebody N.
Apr 16, 2018

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

Explanation:

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)#

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