Ratio of the sum of ##n## terms of two AP series

AI Thread Summary
The discussion focuses on finding the ratio of the 11th terms of two arithmetic progressions (APs) based on their sum formulas. The key expressions provided are the ratios of their sums, given as (S1)n/(S2)n = (7n + 1)/(4n + 27). By substituting n = 21, the ratio of the 11th terms is calculated as (t1)11/(t2)11 = (a1 + 10d1)/(a2 + 10d2) = 148/111. There is a debate about whether the relationship t_n = S_n - S_{n-1} can be used to derive the answer, with some participants expressing uncertainty about the correctness of their calculations. Ultimately, the discussion emphasizes the importance of understanding the ratios and their implications in solving the problem.
brotherbobby
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Homework Statement
The ratio of the sum of ##n## of two APs is ##\dfrac{(7n+1)}{(4n+27)}##. Find the ratio of their ##11^{\text{th}}## terms.
Relevant Equations
1. In an AP with first term ##a## and common difference ##d##, the ##n^{\text{th}}## term is ##t_n=a+(n-1)d##.
2. For the AP in (1) above, the sum of the first ##n## terms is ##S_n = \dfrac{n}{2}\left[ 2a + (n-1)d\right]##
1696926103891.png
Statement of the problem :
I copy and paste the problem as it appears in the text to the right.

Attempt : I must admit I didn't get far, but below is what I did. I use ##\text{MathType}^{\circledR}## hoping am not violating anything.
1696926822683.png


Request : A hint would be very welcome.
 
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You want this
$$\frac{(t_1)_{11}}{(t_2)_{11}}=\frac{a_1+10d_1}{a_2+10d_2}$$
You know this
$$\frac{(S_1)_{n}}{(S_2)_{n}}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}$$
Let ##n=21## in it.
 
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Likes WWGD and pasmith
For an AP we have <br /> S_n = an + \frac{dn(n-1)}{2} = \tfrac12n(nd + (2a - d)) which you seem to have found.

If @martinbn's method were not available, you could say that <br /> \frac{S_n^{(1)}}{S_n^{(2)}} = \frac{nd_1 + (2a_1 - d_1)}{nd_2 + (2a_2 - d_2)} = \frac{C(7n + 1)}{C(4n + 27)} must hold for each n, so that d_1 = 7C, 2a_1 - d_1 = C etc. where C \neq 0 is a common factor which will cancel when you calculate <br /> \frac{a_1 + 10d_1}{a_2 + 10d_2}.
 
martinbn said:
You want this
$$\frac{(t_1)_{11}}{(t_2)_{11}}=\frac{a_1+10d_1}{a_2+10d_2}$$
You know this
$$\frac{(S_1)_{n}}{(S_2)_{n}}=\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}$$
Let ##n=21## in it.
Nice, but don't you mean ##n=11##?
 
WWGD said:
Nice, but don't you mean ##n=11##?
No, you need to cacel a factor of 2.
 
Thank you @martinbn , @pasmith and others. I can solve using @martinbn 's suggestion in post #2.

I am required to find ##\dfrac{(t_1)_{11}}{(t_2)_{11}}=\boxed{\dfrac{a_1+10d_1}{a_2+10d_2}}=?##

I am given however that ##\boxed{\dfrac{(S_1)_{n}}{(S_2)_{n}}=\dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}}=\dfrac{7n+1}{4n+27}##

The two boxed expressions above can be made to match if ##n = 21## in the second result.

Putting ##n = 21##, we obtain ##\mathbf{\dfrac{(t_1)_{11}}{(t_2)_{11}}}= \dfrac{a_1+10d_1}{a_2+10d_2} = \dfrac{7\times 21+1}{4\times 21+27} = \mathbf{\dfrac{148}{111}}##

This is the answer and it matches with the text. ##\checkmark##

Doubt : However, we also know that ##t_n = S_n - S_{n-1}##. Can this fact be used to find the answer?
 
Last edited:
Arguing from my doubt above, I do the following :

1697008824776.png


The answer ##\dfrac{7}{4}## is incorrect.

A hint as to where I have gone wrong above would be very welcome.
 
brotherbobby said:
Arguing from my doubt above, I do the following :

View attachment 333435

The answer ##\dfrac{7}{4}## is incorrect.

A hint as to where I have gone wrong above would be very welcome.
You don't know if ##(S_n)_1=7n+1## and ##(S_n)_2=4n+27##. You just know what their ratio is.
 
martinbn said:
You don't know if ##(S_n)_1=7n+1## and ##(S_n)_2=4n+27##. You just know what their ratio is.
Yes, so I was suspecting. Can we find what is ##(S_n)_1## from the ratio ##\dfrac{(S_n)_1}{(S_n)_2}##?
 
  • #10
brotherbobby said:
Yes, so I was suspecting. Can we find what is ##(S_n)_1## from the ratio ##\dfrac{(S_n)_1}{(S_n)_2}##?

As I showed in my post, (S_n)_i = \tfrac12 n(2a_i + (n-1)d_i) = \begin{cases}<br /> \frac12Cn(7n + 1)&amp; i = 1 \\<br /> \frac12 Cn(4n + 27) &amp; i = 2 \end{cases} for some common factor C \neq 0 which we cannot determine. However, since we only want a ratio, it will cancel.

Then \begin{split}<br /> \frac{(S_n)_1 - (S_{n-1})_1}{(S_n)_2 - (S_{n-1})_2} &amp;= \frac{ n(7n+1) - (n-1)(7n - 6)}{ n(4n + 27) - (n-1)(4n + 23)} \\ <br /> &amp;= \frac{14n - 6}{8n + 23}. \end{split}
 
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