Find the sum of first 6 terms of the G.P. 1, -1/3, 1/3^2, -1/3^3, ........... .
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This is the answer @santhosh
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It is wrong
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Hi ,
1 , - 1/3 , 1/3² , - 1/3³ , ..., is a given G.P
First term = a = 1
Common ratio = r = a2 / a1
r = ( - 1/3 ) / 1
r = - 1/3 < 1
Sum of 6 terms = s6
S6 = [ a ( 1 - r6 ) / ( 1 - r ) ]
= { 1 [ 1 - ( - 1/3 )^6 ]}/ [ 1 - ( - 1/3 ) ]
= [( 3^6 - 1) / 3^6] / [ ( 3 + 1 ) / 3 ]
= ( 3^6 - 1 ) / [ 3^6 ×( 4/3 )]
= ( 3^6 - 1 ) / ( 4 × 3^5 )
I hope this helps you.
:)
1 , - 1/3 , 1/3² , - 1/3³ , ..., is a given G.P
First term = a = 1
Common ratio = r = a2 / a1
r = ( - 1/3 ) / 1
r = - 1/3 < 1
Sum of 6 terms = s6
S6 = [ a ( 1 - r6 ) / ( 1 - r ) ]
= { 1 [ 1 - ( - 1/3 )^6 ]}/ [ 1 - ( - 1/3 ) ]
= [( 3^6 - 1) / 3^6] / [ ( 3 + 1 ) / 3 ]
= ( 3^6 - 1 ) / [ 3^6 ×( 4/3 )]
= ( 3^6 - 1 ) / ( 4 × 3^5 )
I hope this helps you.
:)
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