Question

Super A.S problem :
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Two Different A.S ,The ratio between the two sums from n terms of each is 3n : n+2 ,Prove that the third term of the first one equals the seventh term of the second one ....Then prove that the ratio between T5 of the first sequence and T10 of the second one is 9 : 7
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Answer 1

Not sure how to do ... but this is the beginning
\[ \frac{2a_1 + (n -1) d_1}{2a_2 + (n -1) d_2} = \frac{3n}{n+2} \]\[
(2a_1 + (n -1) d_1)(n+2) = (2a_2 + (n -1) d_2)3n \]

Answer 2

Does that relation hold for any number of terms or for particular number of term?

Answer 3

I believe experimentX is on the right track. I will use different symbols just to avoid confusion. so lets represents the two A.S's as:\[a_n=a_1+(n-1)d_a\]and:\[b_n=b_1+(n-1)d_b\]therefore:\[S_n^a=\frac{n}{2}(2a_1+(n-1)d_a)\]and:\[S_n^b=\frac{n}{2}(2b_1+(n-1)d_b)\]now, using the first ratio from your question, we get:\[\frac{S_n^a}{S_n^b}=\frac{3n}{n+2}\]therefore:\[\frac{\frac{n}{2}(2a_1+(n-1)d_a)}{\frac{n}{2}(2b_1+(n-1)d_b)}=\frac{3n}{n+2}\]the \(\displaystyle\frac{n}{2}\) terms cancel out ad cross-multiplying we get:\[\begin{align}
(n+2)(2a_1+(n-1)d_a)&=3n(2b_1+(n-1)d_b)\\
2a_1n+4a_1+n(n-1)d_a+2(n-1)d_a&=6b_1n+3n(n-1)d_b\\
d_an^2+(2a_1+d_a)n+(4a_1-2d_a)&=3d_bn^2+(6b_1-3d_b)n
\end{align}\]and since this holds for ANY \(n\), we can conclude that:\[d_a=3d_b\tag{1}\]\[2a_a+d_a=6b_1-3d_b\tag{2}\]\[4a_1-2d_a=0\tag{3}\]you should be able to use these three equations to prove the required identities.

Answer 4

@asnaseer :Thats awesome ,ty man
@experimentX :ty bro

Answer 5

yw - thanks to @experimentX for the insight! :)