On the 20th day, you would reach 880.
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Sum(AP) = ½n(2a + (n-1)d)
here:
Sum(AP) = 880
a (first term) = 25
d (common difference) = 2
n (number of terms) = unknown
Plugging in the values gives:
880 = ½n(2×25 + (n - 1) × 2)
→ 880 = n(25 + n - 1)
→ 880 = 24n + n²
→ n² + 24n - 880 = 0
→ (n - 20)(n + 44) = 0
→ n = 20 or -44
You cannot have -44 days, so it must be on the 20th day when he puts 25 + (20 - 1)* 2 = 63 in that he reaches a total of 880.
The sum of an AP can also be expressed as ½ × n × (first + last) since last = first + (n - 1) × difference.
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