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 \(\displaystyle \mbox{ra}\mbox{tio}\) 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
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:
\(\displaystyle \dfrac{a\, +\, \frac{(n\, -\, 1)\,d}{2}}{A\, +\, \frac{(n\, -\, 1)\, D}{2}}\, =\, \dfrac{7n\, +\, 1}{4n\, +\, 27}\)
\(\displaystyle \dfrac{a\, +\, \frac{16}{2}\,d}{A\, +\, \frac{16}{2}\, D}\, =\, \dfrac{7(17)\, +\, 1}{4(17)\, +\, 27}\)
\(\displaystyle \dfrac{a\, +\, 8d}{A\, +\, 8D}\, =\, \dfrac{120}{95}\)
\(\displaystyle \dfrac{t_9}{T_9}\, =\, \dfrac{24}{19}\)
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:
\(\displaystyle \dfrac{(a_9)_1}{(a_9)_2}\, =\, \dfrac{7x}{4x}\, =\, \dfrac{7}{4}\)
Note: I am assuming that "t
9" and "T
9" in the first method refer to the ninth terms of S
1 and S
2, respectively.\)