Question

find the least value of n such that 2+5++8+11+.......upto n trrms >200
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Answer 1

Answer:

2+5+8+11+13+15+17+19+21+23+25+27+29=215

which is the least value on n can have

✔️✔️✔️ therefore n=13✔️✔️✔️

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]Sum=

2

n

[2a+(n−1)d]

Where,

a = first term = 2

d = common difference = 3

Therefore,

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

2

n

[2(2)+(n−1)(3)]

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

2

n

[4+3n−3]

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

2

n

[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