2. Find the sum of the first 27 terms of the geometric series
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GEOMETRIC PROGRESSION :
A sequence a1, a2, a3 ………….. an Is called a geometric sequence if a(n+1) = anr, where r is a non zero constant. Here, a1 is the first term and the constant r is called common ratio. The sequence is also called a geometric progression(G.P)
a1 = a , r = a(ⁿ+1)/ aⁿ
General term of a geometric sequence is
tn = arⁿ-1
General form of G.P is a, a r , a r ²,.........
Sum of n terms in GP ,Sn = a( rⁿ - 1 ) / ( r - 1) [ r> 1] or sn = a(1- rⁿ)⁄(1 - r) [r<1]
SOLUTION :
GIVEN : a1= 1/9 ,a2 = 1/27 and n = 27
r = a(ⁿ+1)/ aⁿ
r = (1/27)/(1/9)
r = (1/27) x (1/9)
r = 1/3
Sum of n terms in GP ,sn = a(1- rⁿ)⁄(1 - r) [r < 1]
S₂₇ = (1/9) [1- (1/3)^27]/[1-(1/3)]
S₂₇ = (1/9) [1- (1/3)^27]/[2/3]
S₂₇= (1/9) x (3/2) [1- (1/3)^27]
S₂₇ = (1/6) [1 - (1/3)^27]
Hence , the sum of 27 terms is (1/6) [1 - (1/3)^27]
HOPE THIS WILL HELP YOU...
A sequence a1, a2, a3 ………….. an Is called a geometric sequence if a(n+1) = anr, where r is a non zero constant. Here, a1 is the first term and the constant r is called common ratio. The sequence is also called a geometric progression(G.P)
a1 = a , r = a(ⁿ+1)/ aⁿ
General term of a geometric sequence is
tn = arⁿ-1
General form of G.P is a, a r , a r ²,.........
Sum of n terms in GP ,Sn = a( rⁿ - 1 ) / ( r - 1) [ r> 1] or sn = a(1- rⁿ)⁄(1 - r) [r<1]
SOLUTION :
GIVEN : a1= 1/9 ,a2 = 1/27 and n = 27
r = a(ⁿ+1)/ aⁿ
r = (1/27)/(1/9)
r = (1/27) x (1/9)
r = 1/3
Sum of n terms in GP ,sn = a(1- rⁿ)⁄(1 - r) [r < 1]
S₂₇ = (1/9) [1- (1/3)^27]/[1-(1/3)]
S₂₇ = (1/9) [1- (1/3)^27]/[2/3]
S₂₇= (1/9) x (3/2) [1- (1/3)^27]
S₂₇ = (1/6) [1 - (1/3)^27]
Hence , the sum of 27 terms is (1/6) [1 - (1/3)^27]
HOPE THIS WILL HELP YOU...
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Solution :
1/9 , 1/27 , 1/81 ,...., is given G.P
First term = a = a1 = 1/9
Common ratio ( r ) = a2/a1
r = ( 1/27 )/( 1/9 )
=> r = 1/3 < 1
Sum of n terms = An
Sn = a[ ( 1 - rⁿ )/( 1 - r ) ]
Here , n = 27 ,
S27 = ( 1/9 ) [ (1-1/3ⁿ )/( 1 - 1/3 )]
= ( 1/9 ) [ ( 1 - 1/3ⁿ ) /( 2/3 ) ]
Sn = ( 1/6 ) ( 1 - 1/3ⁿ )
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