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if the ratio of sum of the first m and n terms of an A.P is m² : n², show that the ratio of its mth and nth terms is (2m-1) : (2n-1

SARAHann  Mar 25, 2017
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 #1
avatar+310 
0

You dont even seem to try and you are pretty much spamming the questions list but ill help you:

Sm=(a+(a+d*(m-1)))*m/2

Sn=(a+(a+d*(n-1)))*n/2

Sm/Sn=(m/n)*(2a+d(m-1))/(2a+d(n-1))=m2/n2

 

Therefore, (2a+d(m-1))/(2a+d(n-1))=m/n

 

What you need to do is to show m=n or 2a=d, then ill show you the answer.

 

Atleast try to do something

Ehrlich  Mar 25, 2017
 #2
avatar+302 
+1

no offence but i'd like it if one of the moderators answered me ok?

SARAHann  Mar 25, 2017
 #3
avatar+75344 
+2

Let a1  be the first term, am be the mth term, an be the nth term and d be the common difference between terms

 

The sum of the first m terms =

m^2 =  m/2 * (a1  + am)  → 2m  =  a1 + am   (1)

 

The sum of the first n terms  =      

n^2  =  n/2 * (a1 + an)  →  2n  =  a1  + an    (2)

 

Subtract  (2)  from (1)

2m - 2n = am - an     (3)  

 

And the  mth term is given by 

am =  a1 + d (m - 1)  →  am  = a1 + dm - d   (4)

 

And the nth term is given by

an = a1 + d(n - 1) →  an  = a1 + dn - d     (5)

 

Subtract  (5) from (4)

am - an   =  dm - dn     (6)

 

Set (3)  and (6)   equal to solve for d

 

2m - 2n   =  dm - dn

2(m - n)   =  d (m - n)

2 = d

 

Rearrange (1)   and solve for a1

am   = a1 + 2m - 2

2m - a1 = a1 + 2m - 2

2m = 2a1 + 2m - 2

0 = 2a1 - 2

2 = 2a1         divide by 2

1  = a1

 

Using (4)  and (5)  and subbing for a1  and  d

am  =  1 + 2m - 2 →  am  = 2m - 1

an =   1   + 2n - 2  →  an  = 2n - 1

 

So.....the ratio of the mth term  to the nth term   =

 

am : an  =    2m - 1   :  2n - 1 

 

 

cool cool cool

CPhill  Mar 26, 2017
 #4
avatar+89835 
+2

Sara I just answered this (on an earlier post) and now I see that CPhill has answered it too.

 

It is ok to make a new post requestingonce again that your question get answered, it is ok to say that you are not happy with your original answer/s,  but please include a link to the original question.

You do not need 2 moderators to answer your question, not both at the same time anyway!

Melody  Mar 26, 2017
 #5
avatar+302 
+2

oh i'm sorry for the inconvience Melody. thank you for caring. and also for answering my question. it's just that i am a little too weak in math and somehow i'm not able to comprehend things a lot,forgive me. but thank you very much for your patience and i do know i was'nt very polite. and yes i'll put the link to the original thread next time. Thank you. smiley

SARAHann  Mar 26, 2017

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