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

1. ## 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.
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=>Second method
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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!

2. Originally Posted by navneet9431
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?"
In this case the answer is 24/19.

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?
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.

3. Originally Posted by stapel
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.

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

4. Originally Posted by navneet9431
I have made changes in my post and I will keep this in mind while writing anything here in future.
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.)

Question: If the $\mbox{ra}\mbox{tio}$ 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 $\mbox{ra}\mbox{tio} of their 9th terms? 1st solution method: Let S1 have first term "a" and common difference "d". Let S2 have first term "A" and common difference "D". Then: [tex]S_1\, :\, S_2\, =\, (7n\, +\, 1)\, :\, (4n\, +\, 27)$

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

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

$\dfrac{t_9}{T_9}\, =\, \dfrac{a\, +\, 8d}{A\, +\, 8D}$

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

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

Then:

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

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

$\dfrac{a\, +\, 8d}{A\, +\, 8D}\, =\, \dfrac{120}{95}$

$\dfrac{t_9}{T_9}\, =\, \dfrac{24}{19}$

2nd solution method:

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

Then:

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

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

. . . . .$=\, 64x\, -\, 57x\, =\, 7x$

Also:

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

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

. . . . .$=\, 63x\, -\, 59x\, =\, 4x$

Therefore:

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

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