Question

how many terms of the progression 3,6,9,10


must be taken at the least to have a sum not less than 2000?​

Answer 1

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

2000 = n/2 (2(3)+(n-1)3)

4000 = n(6+3n-3)

4000= 3n²+3n

3n²+3n-4000 =0

n = 36