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

#1**0 **

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

S_{m}=(a+(a+d*(m-1)))*m/2

S_{n}=(a+(a+d*(n-1)))*n/2

S_{m}/S_{n}=(m/n)*(2a+d(m-1))/(2a+d(n-1))=m^{2}/n^{2}

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

#3**+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

CPhill
Mar 26, 2017

#4**+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**+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.

SARAHann
Mar 26, 2017