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

→ 2sum = n(2a + (n-1)d)

→ 2sum = 2an + dn² - dn

→ dn² +(2a - d)n - 2sum = 0

a = 3 (first term)

d = 4 (common difference)

sum = 820

→ 4n² + (2×3 - 4)n - 2 × 820 = 0

→ 4n² + 2n - 2 × 820 = 0

→ 2n² + n - 820 = 0

→ (2n + 41)(n - 20) = 0

As n must be positive, the (2n + 41) gives a negative value for n so cannot be a solution, leaving

n - 20 = 0

→ n = 20

There are 20 terms in the AP.

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