Find the sum of all natural numbers between 25 and 210 which are either divisible by 3 or divisible by 4? Please let me know ​I will mark you Brainliest

Let S be the sum of the integers 25-210:

S = 25 + 26 + 27 + … + 208 + 209 + 210

Let S₃, S₄, and S₁₂ denote the sums of the integers in S that are multiples of 3, 4, or 12, respectively. We'll also count how many terms each sum involves; it'll be useful later.

S₃ = 27 + 30 + 33 + … + 204 + 207 + 210 … … … (62 terms)

S₃ = 3 (9 + 10 + 11 + … + 68 + 69 + 70)

S₄ = 28 + 32 + 36 + … + 200 + 204 + 208 … … … (46 terms)

S₄ = 4 (7 + 8 + 9 + … + 50 + 51 + 52)

S₁₂ = 36 + 48 + 60 + … + 180 + 192 + 204 … … … (15 terms)

S₁₂ = 12 (3 + 4 + 5 + … + 15 + 16 + 17)

Let's look at S₃ :

S₃ = 3 (9 + 10 + 11 + … + 68 + 69 + 70)

By reversing the order of the sum, we get

S₃* = 3 (70 + 69 + 68 + … + 11 + 10 + 9)

Of course S₃ = S₃*, I'm just calling it something else temporarily. Notice that every term in the same position of either sum adds up the same number.

9 + 70 = 79

10 + 69 = 79

11 + 68 = 79

and so on. Then

S₃ + S₃* = 3 (79 + 79 + 79 + … + 79 + 79 + 79)

or

2S₃ = 3 × 62 × 79   ==>   S₃ = 7,347

We can compute the other two sums in the same way.

S₄ = 4 (7 + 8 + 9 + … + 50 + 51 + 52)

S₄* = 4 (52 + 51 + 50 + … + 9 + 8 + 7)

==>   2S₄ = 4 × 46 × 59  ==>   S₄ = 5,428

S₁₂ = 12 (3 + 4 + 5 + … + 15 + 16 + 17)

S₁₂* = 12 (17 + 16 +15 + … + 5 + 4 + 3)

==>   2S₁₂ = 12 × 15 × 20   ==>   S₁₂ = 1,800

Then the sum you want is

S₃ + S₄ - S₁₂ = 10,975

We subtract S₁₂ because each of its terms is counted twice (once in S₃ and again in S₄).

step-by-step explanation:

4227473658779875563886538705958641728563470543621287463824132573847969p05749365273847906970680574936214

a. 12.5 kg

step-by-step explanation:

f = ma

so

m = f/a = (2500 n)/(200 m/s²) = 12.5 kg

1 n = 1 kg·m/s²