Question

The least value of n that 1+3+5+7+........n terms ≥500 is

Answer 1

1+3+5+7+...
=n(1+(2n-1))/2=n²

n²≥500, so, minimum value of n=23

Answer 2

Answer:

23

Step-by-step explanation:

Here,

a (first term) = 1

d (common difference) = 2

Sn = 500Substitute the above values in the formula Sn=n2[2a+(n−1)d]. So,

500=n2[2(1)+(n−1)(2)]500=n2[2+2n−2]500=n2(2n)500=n2n=500‾‾‾‾√n=22.36

Thus, the least value which n can take is 23

Therefore, n = 23