Find the least value of n for which 3+6+9+...+ to n terms exceeds 500 .
a-18
b-20
c-21
d-22
e-none of above
pls explain
a-18
b-20
c-21
d-22
e-none of above
pls explain
Is the answer A?
Sum of n numbers
Sn = (n/2) (2a1 + (n-1)d)
d = difference between terms = 6-3 = 3
n = number of terms
a1 = first term = 3
Sn = sum > 500
Sn = (n/2) (2a1 + (n-1)d)
500 > (n/2) (2a1 + (n-1)d)
500 > (n/2) (2 X 3 + (n-1)3)
500 > (n/2) (6 + 3n - 3)
500 > (n/2) (3 + 3n)
Substitue 17 and 18
When you substitue 17 (number of terms), the sum will be less than 500
When you substitue 18 (number of terms), the sum will be greater than 500
