Given that n AM's are inserted between two sets of numbers a, 2b and 2a, b , where a and b are rational numbers. Suppose further that mth mean between these sets of number is same then the ratio of a:b equals
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OpenStudy (anonymous):
@sathiprakash
OpenStudy (anonymous):
im sorry dude.....hope other helps you...im not in a mood now thanks
OpenStudy (anonymous):
@gerryliyana
OpenStudy (anonymous):
@mathslover
OpenStudy (anonymous):
@BostonBlue
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OpenStudy (anonymous):
@Mashy
mathslover (mathslover):
See,
I can give you a hint as I have to go for dinner
mathslover (mathslover):
Can you please wait for 15 minutes? I will help you then ...
OpenStudy (anonymous):
nevermind take your time
OpenStudy (anonymous):
@mathslover
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OpenStudy (anonymous):
@BostonBlue got any idea
mathslover (mathslover):
try to go like this :
2b = a + nd
b = 2a + nd'
\(\cfrac{2b-a}{n} = d\) ------ (1) and \(\cfrac{b-2a}{n} = d'\) ----------(2)
T_(m+1) = a + md...
T' _ (m+1) = 2a + m/n (b-2a)
mathslover (mathslover):
@Meinme did you get what I meant?
mathslover (mathslover):
@Meinme are you there?
OpenStudy (anonymous):
yeah
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OpenStudy (anonymous):
i did it same but not getting the answer ILL check it once and come back