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Let N be any non-zero integer. Then 58*N divided by N will always give 58 as their quotient.
t(n) = (5n3 - 30n2 + 85n - 48)/6 , n = 1, 2, 3, ...
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
t(n) = 3*2n-1 Or just 3*2n
You can either do this by trial-and error, or solve the equation n + (n + 2) + (n + 4) = 48.