AP...

From our observations,
1,3,5,.... is an AP

Again, 2, 4, 6,.......is also an AP.

Use the formula for sum of n terms of an AP to evaluate each of them.......

1 + 3 + 5 + ..........+ (2n - 1)2 + 4 + 6 + ......2n

For numerator,
No. of terms = n
nth term, or last term = 2n - 1
So, S₁ = n2 (1 + 2n -1 ) = n²

you can directly write this if you know sum of consecutive odd terms is equal to square of the number of terms.

For denominator,
No. of terms = n
nth term, or last term = 2n
So, S₂ = n2 (2 + 2n) = 2n(n +1)2 = n(n + 1)

Now, S₁S₂ = n²n(n + 1) = nn + 1 = 115116

→ 116n = 115n + 115
→ n = 115