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Anurup replied | 02/05/2016

a=1st term ; r=common ratio
Sum=a/(r-1), if r>1
Sum=a/(1-r), if r<1

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Ashutosh replied | 13/05/2016

it is a/(1-r) provided |r|<1 otherwise series can't converge.

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Seerat replied | 28/04/2016

yes relationship between AM,HM and GM can be shown by this equatiion
AM x HM =GM^2

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Ashutosh replied | 17/05/2016

GM is actually the GM of AM and HM. i.e. GM^2 = AM*HM
also AM>GM>HM always.

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Ashutosh Ranjan replied | 27/10/2016

1:6

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Ashutosh Ranjan replied | 27/10/2016

Sn= n/2[2a+(n-1)d
S1=1/2[2a+(1-1)d]
S1= a
Now,
S4=4/2[2a+(4-1)d]
S4= 4a+6d
S1/ s4= 1/10
a/4a+6d= 1/10
a=d
S3= 3/2[2a+(3-1)d]
S3= 3/2[2a+2a], because a=d
S3=6a
Hence,
S1/s3 = a/6a
S1/s3 = 1/6

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Avinash replied | 27/04/2016

let the numbers be a-3d, a-d, a+d, a+3d,
given their sum is 20.
a-3d+a-d+a+d+a+3d=20
after solving, a=5....(1)
we know
(a-3d)(a-3d)/(a-d)(a+d) = 2/3...(2)
from (1), substitute the value of a in (2)
by solving we will get d=+1 or -1

after substituting the values of a & d, we get the numbers as 2,4,6,8

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John replied | 09/05/2016

2,4,6,8- found out by mere guessing.

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