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Mathematics 16 Online
OpenStudy (anonymous):

If the common difference in an arithmetic is twice the first term, show that Sn/Sm=n^2/m^2

terenzreignz (terenzreignz):

Just to be clear, in an arithmetic progression, the sum until the nth term is given by... \[\huge S_n = \frac{n(a_1+a_n)}{2}\]

OpenStudy (anonymous):

ok

terenzreignz (terenzreignz):

So, what have you tried so far?

OpenStudy (anonymous):

i don't know how to start

terenzreignz (terenzreignz):

Okay, it would help to have a formula for ... \[\huge a_n = ?\]

terenzreignz (terenzreignz):

In general, when given an arithmetic progression, \[\huge a_n = a_1+(n-1)d\]where d is the common difference. We have an expression for the common difference, what is it?

OpenStudy (anonymous):

its twice the first term

terenzreignz (terenzreignz):

And the first term is \(\large a_1\) Therefore... \[\huge d = 2a_1\] Substituting this... \[\huge a_n =a_1+(n-1)(2a_1) \] Please simplify this.

OpenStudy (anonymous):

can u just give me an answer

OpenStudy (anonymous):

@Mertsj

OpenStudy (mertsj):

Nobody can just give you an answer because the question says to SHOW so why don't you go ahead and simplify the expression that terenz posted?

OpenStudy (anonymous):

fine a1+(n-1)d

OpenStudy (anonymous):

how does this help

OpenStudy (mertsj):

He gave you the expression where he substituted for d per the problem instructions. That's the one you need to simplify

OpenStudy (mertsj):

And what's with the hostile attitude toward the people who are trying to help you.

OpenStudy (mertsj):

And I don't know how it will help...guess we'll find out as the problem progresses.

OpenStudy (mertsj):

The asker has disappeared.

OpenStudy (anonymous):

well thats annoying.. had the answer for him.. aaai

OpenStudy (mertsj):

What is Sm?

OpenStudy (anonymous):

i am back @Mertsj

OpenStudy (anonymous):

@Mertsj Sm means the sum of all terms till the mth term.

OpenStudy (anonymous):

can u help me lucky

OpenStudy (anonymous):

I'll try to simplify the problem. Not guaranteed that i'll make it baby-ish. Now As the main replier has correctly said - Nth term = a + (n-1)2a = a + 2an -2a => 2an - a => a(2n-1) Mth term = a + (m-1)2a => 2am - a = a(2m-1) Now sum till the Nth = N/2 ( a + a(2n-1)) Sum till the Mth = M/2 ( a + a(2m-1) => So Sm/Sn = N/2(a + 2an - a)/ M/2(a + 2am - a) => 2 is cancelled, a-a = 0 => Sm/Sn = n(2an)/m(2am) Simplify. :)

OpenStudy (anonymous):

k thanks

OpenStudy (anonymous):

i got Sn/Sm=n^2/m^2

OpenStudy (mertsj):

That's what the directions said to show. Good job!!

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