the sum of the series 243,81,27,...to 8 terms is
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Answered by
34
the given series is in geometric progession
where a=243
r=1/3
n=8
their sum is a(r^n-1)/(r-1)
243((1/3)^8-1)/(-2/3)
364.4
I hope this will help u ;)
where a=243
r=1/3
n=8
their sum is a(r^n-1)/(r-1)
243((1/3)^8-1)/(-2/3)
364.4
I hope this will help u ;)
ak0071:
tghanks
welcome
Answered by
38
Use the formula of G.P
Here every next term is a multiple of 1/3
Sn= a(1-r^n)/1-r
Where a=243
n=8
r=1/3
Substitute in the formula
Sn=243(1-[1/3]^8)/(1-1/3)
=243(1-(1/6561) / (2/3)
=243[6560/6561] / (2/3)
=(243×6560×3)/6561×2
=4782240/13122]
=364.444
~364
Here every next term is a multiple of 1/3
Sn= a(1-r^n)/1-r
Where a=243
n=8
r=1/3
Substitute in the formula
Sn=243(1-[1/3]^8)/(1-1/3)
=243(1-(1/6561) / (2/3)
=243[6560/6561] / (2/3)
=(243×6560×3)/6561×2
=4782240/13122]
=364.444
~364
Hope it helps..!!!! Comment in case of any doubt
the sum of the series 1/ \sqrt{3} +1+3/ \sqrt{3} + ... to 18 terms
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Can you give the link of the question plzz
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