In an infinite geometric progression,the sum of first two terms is 6 and every term is four times the sum of all the terms that follow it.find 1. The geometric progression 2.its sum to infinity. Plz help
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Geometric series is a, a r , a r², ... a = first term and common ratio is r
Tn = a r^(n-1) Sn = a (1 - r^n) / (1-r)
Sum to infinity = a / (1-r) if r < 1
a + a r = 6 a ( 1 + r ) = 6 -- equation 1
a = 4 * [ a r + a r² + ... ] = 4 a r [ 1 + r + r² ..] = 4 a r / (1 -r )
SO, 1 - r = 4 r => r = 0.2 or 1/5
a = 6/1.2 = 5
G. Progression is 5, 1/5, 1/25, ....
Sum to infinity: a / (1-r) = 5 / 0.8 = 6.25
Tn = a r^(n-1) Sn = a (1 - r^n) / (1-r)
Sum to infinity = a / (1-r) if r < 1
a + a r = 6 a ( 1 + r ) = 6 -- equation 1
a = 4 * [ a r + a r² + ... ] = 4 a r [ 1 + r + r² ..] = 4 a r / (1 -r )
SO, 1 - r = 4 r => r = 0.2 or 1/5
a = 6/1.2 = 5
G. Progression is 5, 1/5, 1/25, ....
Sum to infinity: a / (1-r) = 5 / 0.8 = 6.25
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