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

[brotherbobby]

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 $S_n = \dfrac{n}{2}\left[ 2a + (n-1)d\right]$

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.

Request : A hint would be very welcome.

 

[martinbn]

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.

 

[pasmith]

For an AP we have $$S_n = an + \frac{dn(n-1)}{2} = \tfrac12n(nd + (2a - d))$$ which you seem to have found.

If @martinbn's method were not available, you could say that $$\frac{S_n^{(1)}}{S_n^{(2)}} = \frac{nd_1 + (2a_1 - d_1)}{nd_2 + (2a_2 - d_2)} = \frac{C(7n + 1)}{C(4n + 27)}$$ 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 $$\frac{a_1 + 10d_1}{a_2 + 10d_2}.$$

 

[WWGD]

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 $n=11$?

 

[martinbn]

Nice, but don't you mean $n=11$?

No, you need to cacel a factor of 2.

 

[brotherbobby]

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?

 

[brotherbobby]

Arguing from my doubt above, I do the following :

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

A hint as to where I have gone wrong above would be very welcome.

 

[martinbn]

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.

 

[brotherbobby]

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 $\dfrac{(S_n)_1}{(S_n)_2}$?

 

[pasmith]

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, $$(S_n)_i = \tfrac12 n(2a_i + (n-1)d_i) = \begin{cases} \frac12Cn(7n + 1)& i = 1 \\ \frac12 Cn(4n + 27) & i = 2 \end{cases}$$ for some common factor $C \neq 0$ which we cannot determine. However, since we only want a ratio, it will cancel.

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