Question

How many terms of the AP 3, 7,11.........must be taken their sum is 300?​

Answer 1

a= 3

d=7-3=11-7=4

so ,

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

300=n/2(2×3+(n-1)×4)

600=6n+4n2-4

4n2+6n-604=0

2n2+3n-302=0

Now by using quadratic formula you can solve it

i.e. (-b+-√(b×b-4ac))/2a

where a=2

b=3

and c=-302

Hope it helped.....