How many terms of the progression 3,6,9,12 ... must be taken at the least to have a sum not less than 2000?

You need at least 37 terms to have a sum of 2000 or more

Step-by-step explanation:

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S (n) : =3+6+9+12+⋯+3n=3⋅n (n+1) 2

This is approximately 32⋅n2.

It should be around 2000, that is n2 should be around 2/3⋅2000≈1334. Using calculator, its square root is ≈36.52 and if you sustitue n=36 to S (n) gives S (36) = 1998.

So since 36 is not enough you need 37 terms to have a sum of 2000 or more