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Yes.

There are 2 ways to interpret your question:

1. You are adding 5 + 10 + 15 + 20 + 25 (= 75), and then adding another 5 + 10 + 15 + 20 + 25 (= another 75 making 150 so far), and so on

In this case 10,000 ÷ 75 = 133⅓ times

133 times would get a total of 9,975

134 times would get a total of 10,050

2. You are adding 5 + 10 + 15 + 20 + 25 + 30 + ... each time adding 5 more than last time.

In this case you have an AP with first term 5 and common difference 5.

Sum = (n/2)(2a + (n-1)d)

→ 10,000 = (n/2)(2×5 + (n-1)×5)

→ 20,000 = n(10 + 5n - 5)

→ 20,000 = 5n² + 5n

→ n² + n = 4,000

→ n² + n - 4,000 = 0

→ n = (-1 ±√(1+4×4,000)/2

→ n = (-1 ±√16,001)/2

As n must be positive, we can ignore the negative square root to get:

n = (√16,001 - 1)/2 ≈ 62.7 times

62 times would get a total of 9,765,

63 times would get a total of 10,080.

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