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Aamir 17/05/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition)

Which tuition is better, online or offline?

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Shefali replied | 26/06/2016

Depending upon how much you are comfortable with technology you can choose mode of teaching. If well versed with technology, online is better option coz we can get paid well but if not start with offline and move ahead with online. Personally I would always opt for offline for personal touch to students.

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Abdul 12/05/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition)

DO TEACHERS CHARGE FOR ONLINE TEACHING

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Shefali replied | 26/06/2016

Yes. You pay for their services towards teaching. Even rates would vary.

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Prasanth Kumar replied | 15/10/2016

Yes.we charge fee depending on the time period for which a student needs our guidance

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U.K. 01/05/2016 in  Class IX-X Tuition, Mathematics(Class IX-X Tuition)

what is the least remainder when 17 power 30 is divided by 5

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Ashutosh replied | 06/05/2016

the unit digit of 17^30 is 9. hence the remainder is 4.

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Navin replied | 17/05/2016

The answer to this question is very simple. its 4.
Unit digit for 17^n, will be 7,9,3,1,7,9,3,1... so on. As you see it is getting same numbers after certain period, so if n=30, the unit digit will be 9.
And remainder after dividing the value by 5 will be "4". :) :)

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Meenakshi replied | 03/11/2016

If d (common difference) is same for different values of n , then

the given list of numbers form an AP .

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Sourabh replied | 04/11/2016

As all the answers say
If the difference is constant its in AP.
I will like to add one point that nothing but just practice can help you get through this problem.

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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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