Let x and d be the first term and common difference respectively of the AP, respectively.
Then,

Let S_m and S_n be the sum of the first m and first n terms of the A.P. respectively. Let, a be the first terms and d be a common difference.



⇒[2a+(m−1)d]
n =[2a(n−1)d]m
⇒2an+mnd−nd=2am+mnd−md
⇒md−nd=2am−2an
⇒(m−n)d=2a(m−n)
⇒d=2a
Now, the ratio of its mth and nth terms is


Thus, the ratio of its mth and nth terms is 2m – 1 : 2n – 1.

The given AP is 8, 10, 12, ....

So,

First term =a = 8

Common difference = d = 10-8 =2

We know that n^th term of an AP, a_n = a + (n - 1)

60^th term of the given AP = a₆₀ = 8 +( 60-1) x  2 = 8 + 59 x 2 = 8 + 118 = 126

Therefore, the 60^th term of the given AP is 126

It is given that the AP has a total of 60 terms. So, in order to find sum of last n terms. we take

First term, A = 126
Common difference, D = -2
Now,

Let the numbers be (a -3d),(a-d), (a+d) and (a+3d)

∴ (a-3d) + (a-d) +(a+ d) + (a +3d) = 32

⇒ 4a = 32

a = 8

$$\begin{gathered} Also, \frac{(a-3d)(a+ 3d)}{(a-d)(a+d)} =\frac{7}{15} \\ \Rightarrow 15a^2 -135d^2 = 7a^2 -7d^2 \\ \Rightarrow 8a^2 = 128d^2 \\ {d}^2 = \frac{8a^2}{128} = \frac{8 x 8x8}{128} \\ {d}^2 =4 \\ d =\pm 2 \end{gathered}$$

If d =2 numbers are : 2, 6,10, 14
If d = -2 numers are 14,10,16,2