why am i getting different answers by different methods to same question? (sequence)

I have tried solving this question using two different methods- "If the ratio of the sum of the first n terms of two APs is (7n+1) : (4n+27), then what is the ratio of their 9th terms?"
=>The first method
In this case the answer is 24/19.


=>Second method


In the second method i used this formula an=S_n-S(n-1) So For the first A.P. Sum of "n" terms =(7n+1)*x [using the formula] So, a9=S_9-S(8) We will finally get a_9=7x Similarly ,for the second A.P. we will have A_9=4x So,Ratio of the ninth term of both the A.P is 7:4(Check attachment for more!)

I want to know why am i getting two different answers?Are both the methods correct?IF NOT,then please ELABORATE why?

I will be thankful for help!

navneet9431 said:

I have tried solving this question using two different methods- "If the ratio of the sum of the first n terms of two APs is (7n+1) : (4n+27), then what is the ratio of their 9th terms?"
=>The first method(click here- [deleted for security])
In this case the answer is 24/19.

=>Second method(click here- [deleted for security])
In the second method i used this formula an=S_n-S(n-1) So For the first A.P. Sum of "n" terms =(7n+1)*x [using the formula] So, a9=S_9-S(8) We will finally get a_9=7x Similarly ,for the second A.P. we will have A_9=4x So,Ratio of the ninth term of both the A.P is 7:4(Check attachment for more!)

I want to know why am i getting two different answers?Are both the methods correct?IF NOT,then please ELABORATE why?

Click to expand...

Probably the reason nobody has answered is that nobody wants to go to the trouble to retrieve your images (especially when they're hosted on a proscribed site: see warning here) and type them out for you here, before being able to begin responding.

So please reply by uploading the images to the forum server, or else by typing out your work, so that we can see it here. Thank you!

stapel said:

Probably the reason nobody has answered is that nobody wants to go to the trouble to retrieve your images (especially when they're hosted on a proscribed site: see warning here) and type them out for you here, before being able to begin responding.

So please reply by uploading the images to the forum server, or else by typing out your work, so that we can see it here. Thank you!

Click to expand...

Thank you so much!
I have made changes in my post and I will keep this in mind while writing anything here in future.

navneet9431 said:

I have made changes in my post and I will keep this in mind while writing anything here in future.

Click to expand...

In future, please put new content into new posts, rather than adding them as edits to previous posts. This helps ensure that the "conversation" makes sense to readers.

The images are rather small inside your edited post. I'll type them out for you. (You have beautiful handwriting, by the way.)

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Question: If the \(\displaystyle \mbox{ra}\mbox{tio}\) of the sum of the first n terms of two arithmetic progressions (that is, arithmetic sequences), S₁ and S₂, is (7n+1) : (4n+27), then what is the \(\displaystyle \mbox{ra}\mbox{tio} of their 9th terms?\)\(\displaystyle

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1st solution method:


Let S₁ have first term "a" and common difference "d". Let S₂ have first term "A" and common difference "D". Then:

\(\displaystyle S_1\, :\, S_2\, =\, (7n\, +\, 1)\, :\, (4n\, +\, 27)\)

\(\displaystyle \dfrac{S_1}{S_2}\, =\, \dfrac{\left(\frac{n}{2}\right)\, \left[2a\, +\, (n\, -\, 1)\, d\right]}{\left(\frac{n}{2}\right)\, \left[2A\, +\, (n\, -\, 1)\, D\right]}\)

. . . . . . .\(\displaystyle =\, \dfrac{a\, +\, \frac{(n\, -\, 1)\,d}{2}}{A\, +\, \frac{(n\, -\, 1)\, D}{2}}\)

\(\displaystyle \dfrac{t_9}{T_9}\, =\, \dfrac{a\, +\, 8d}{A\, +\, 8D}\)

\(\displaystyle \mbox{From }\, \dfrac{a\, +\, \frac{(n\, -\, 1)\,d}{2}}{A\, +\, \frac{(n\, -\, 1)\, D}{2}}\, \mbox{ and }\, \dfrac{a\, +\, 8d}{A\, +\, 8D}\)

\(\displaystyle \mbox{we get }\, \dfrac{n\, -\, 1}{2}\, =\, 8,\, \mbox{ so }\, n\, =\, 17.\)

Then:

\(\displaystyle \dfrac{a\, +\, \frac{(n\, -\, 1)\,d}{2}}{A\, +\, \frac{(n\, -\, 1)\, D}{2}}\, =\, \dfrac{7n\, +\, 1}{4n\, +\, 27}\)

\(\displaystyle \dfrac{a\, +\, \frac{16}{2}\,d}{A\, +\, \frac{16}{2}\, D}\, =\, \dfrac{7(17)\, +\, 1}{4(17)\, +\, 27}\)

\(\displaystyle \dfrac{a\, +\, 8d}{A\, +\, 8D}\, =\, \dfrac{120}{95}\)

\(\displaystyle \dfrac{t_9}{T_9}\, =\, \dfrac{24}{19}\)

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2nd solution method:


\(\displaystyle \mbox{Let }\, S_{n_1}\, \mbox{ be }\, (7n\, +\, 1)\, x\, \mbox{ and let }\, S_{n_2}\, \mbox{ be }\, (4n\, +\, 27)\, x\)

Then:

\(\displaystyle a_{n_1}\, =\, S_{n_1}\, -\, S_{(n-1)_1}\)

\(\displaystyle (a_9)_1\, =\, \left(S_9\right)_1\, -\, \left(S_8\right)_1\)

. . . . .\(\displaystyle =\, 64x\, -\, 57x\, =\, 7x\)

Also:

\(\displaystyle a_{n_2}\, =\, S_{n_2}\, -\, S_{(n-1)_2}\)

\(\displaystyle (a_9)_2\, =\, \left(S_9\right)_2\, -\, \left(S_8\right)_2\)

. . . . .\(\displaystyle =\, 63x\, -\, 59x\, =\, 4x\)

Therefore:

\(\displaystyle \dfrac{(a_9)_1}{(a_9)_2}\, =\, \dfrac{7x}{4x}\, =\, \dfrac{7}{4}\)

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Note: I am assuming that "t₉" and "T₉" in the first method refer to the ninth terms of S₁ and S₂, respectively.\)