Question

Find the least value of n such that 2+5+8+11+...to n terms >=200

Answer 1

Hope this will help you

Answer 2

Answer:

Least value of n = 12

Step-by-step explanation:

As per the question

The given series

2 + 5 + 8 + 11 + .....

As we can see that the common difference is same and equals to 3.

5 - 2 = 3 = 8 - 5

Therefore,

The given series is in A.P

The sum of n terms of A.P is given by

$Sum = \frac{n}{2}[2a+(n-1)d]$

Where,

a = first term = 2

d = common difference = 3

Therefore,

$Sum = \frac{n}{2}[2(2)+(n-1)(3)]$

$Sum = \frac{n}{2}[4+3n-3]$

$Sum = \frac{n}{2}[3n+1]$

n(3n+1) ≥ 400

3n² + n - 400 ≥ 0

On solving this, we get

n ≥ 11.381

Hence, the sum for greater than 200, the least value of n = 12