How many terms of the progression 3,6,9, 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:

,

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

step-by-step explanation:

to find if any function is an inverse of the other, put the first equation in as x in the next one. then simplify and solve. if at the end you get the statement y = x, then it is an inverse. if you get anything else, it is not.