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
You dont even seem to try and you are pretty much spamming the questions list but ill help you:
What you need to do is to show m=n or 2a=d, then ill show you the answer.
Atleast try to do something
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
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!
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.