Question

How many terms of the series 45+39+33 amount to 192​

Answer 1

Answer:

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

192=n/2(2×45+(n-1)-6)

192=n/2(90-6n+6)

192=n(45-3n+3)

192=48n+3n²

3n²+48n-192=0

3(n²+16n-64)=0

n²+16n-64

solve this and you will get the answer

thank u