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 [tex]
S_n = an + \frac{dn(n-1)}{2} = \tfrac12n(nd + (2a - d))[/tex] which you seem to have found.

If @martinbn's method were not available, you could say that [tex]
\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)}[/tex] must hold for each [itex]n[/itex], so that [itex]d_1 = 7C[/itex], [itex]2a_1 - d_1 = C[/itex] etc. where [itex]C \neq 0[/itex] is a common factor which will cancel when you calculate [tex]
\frac{a_1 + 10d_1}{a_2 + 10d_2}.[/tex]

 

[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, [tex](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}[/tex] for some common factor [itex]C \neq 0[/itex] which we cannot determine. However, since we only want a ratio, it will cancel.

Then [tex]\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}[/tex]