find the least value of n such that 2+5+8+11+....to n terms > 200.
12
Sn = n/2 (2*2+(n-1)*3) > 200
Ans
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To find the least value of n such that the sum of the series exceeds 200, we need to determine the pattern of the series and then use that pattern to calculate the sum.
The given series follows an arithmetic sequence, where each term is obtained by adding a constant difference (d) to the previous term. In this case, the common difference (d) is 3, as we can see that each term increases by 3.
The arithmetic series formula is given by Sn = (n/2)(2a + (n-1)d), where Sn is the sum of the series, n is the number of terms, a is the first term, and d is the common difference.
In this series, the first term (a) is 2 and the common difference (d) is 3. We want to find the least value of n such that the sum (Sn) is greater than 200.
Let's substitute the given values into the formula and solve for n:
Sn > 200
(n/2)(2a + (n-1)d) > 200
(n/2)(2(2) + (n-1)(3)) > 200
(n/2)(4 + 3n - 3) > 200
(n/2)(3n + 1) > 200
3n^2/2 + n/2 > 200
Now, we have a quadratic inequality. Let's solve it:
3n^2 + n > 400
3n^2 + n - 400 > 0
To solve this equation, you can either factorize it or use quadratic formula:
Using quadratic formula:
n = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 3, b = 1, c = -400
n = (-(1) ± √((1)^2 - 4(3)(-400))) / (2(3))
n = (-1 ± √(1 + 4800)) / 6
n = (-1 ± √4801) / 6
Since we are looking for the least value of n, we can ignore the negative value.
n = (√4801 - 1) / 6
n ≈ 15.77 (rounded to two decimal places)
Since n represents the number of terms in the series, it must be a whole number. Therefore, the least value of n is 16 which satisfies the condition of the sum of the series being greater than 200.