1)The nth term of a progression is 6(#-4/3##)^2#. Prove that the progression is a geometric progression(G.P.).Obtain the first term and common ratio. (one more question, pls see below). ?

Question

.2)Find three consecutive term of a G.P. where the sum and product of these three terms are 65 & 3375 respectively.

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Answer 1 · 2018-07-29T11:15:25

1) First term and common ratio of G.P. are # 6 and (-4/3)#
2) Three consecutive terms are # (5,15,45) or (45,15,5)#

1) #n# th term of G.P is # a r^(n-1) :. n-1=2 or n=3#

#:.6(-4/3)^2# is the #3# rd term of G.P of which #a=6# is

the first term and common ratio is #r=-4/3#.

2) Let the three consecutive terms of G.P are

#a/r, a, a r#. Product of three consecutive terms is

#a/r * a* a r=3375 or a^3 =3375 :. a= root(3) 3375# or

#a= 15#. Sum of three consecutive terms is

#a/r + a +a r=65 or a(r+1/r+1)=65# or

#15(r+1/r+1)=65 or (r^2+r+1)/r= 13/3# or

#3 r^2+3 r+3-13r=0 or 3 r^2-10 r+3=0# or

#3 r^2-9 r-r+3=0 or 3r(r-3)-1(r-3)=0#

# r=3 , r=1/3# .When #r=3# three consecutive terms are

# 15/3, 15, 15*3 or (5,15 , 45)# ,when #r=1/3# three consecutive

terms are # 15/(1/3), 15, 15/3 or (45,15, 5)#[Ans]